Two guns and can fire bullets at speeds and respectively. From a point on a horizontal ground, they are fired in all possible directions. The ratio of maximum areas covered by the bullets fired by the two guns, on the ground is :
1
1:02
2
1:04
3
1:08
4
1:16
Official Solution
Correct Option: (4)
02
PYQ 2021
medium
physicsID: jee-main
A player kicks a football with an initial speed of 25 ms at an angle of 45 from the ground. What are the maximum height and the time taken by the football to reach at the highest point during motion? (Take g = 10 ms )
1
h = 10 m T = 2.5 s
2
h = 15.625 m T = 3.54 s
3
h = 15.625 m T = 1.77 s
4
h = 3.54 m T = 0.125 s
Official Solution
Correct Option: (3)
Step 1: Understanding the Question:
We are given the initial velocity and angle of projection of a football. We need to find the maximum height it reaches and the time it takes to get there. This is a standard projectile motion problem. Step 2: Key Formula or Approach:
For a projectile launched with initial speed at an angle with the horizontal:
1. Time to reach maximum height:
2. Maximum height: Step 3: Detailed Explanation:
Given values:
Initial speed,
Angle of projection,
Acceleration due to gravity, First, let's calculate the time taken to reach the highest point (T).
We know that .
Rounding to two decimal places, . Next, let's calculate the maximum height ( ).
We can also write this as , where is the initial vertical velocity.
To simplify the fraction, divide numerator and denominator by 5: . Step 4: Final Answer:
The maximum height reached is 15.625 m, and the time taken to reach it is 1.77 s. This corresponds to option (C).
03
PYQ 2022
hard
physicsID: jee-main
A ball of mass m is thrown vertically upward. Another ball of mass 2 m is thrown at an angle \theta with the vertical. Both the balls stay in air for the same period of time. The ratio of the heights attained by the two balls respectively is 1/x. The value of x is _______.
Official Solution
Correct Option: (1)
To solve this problem, we need to compare the heights attained by two balls using their respective flight properties. When the ball is thrown vertically, its equations of motion determine its maximum height. Meanwhile, for the ball thrown at an angle, the vertical component of its velocity determines its time in the air and maximum height. The problem states both balls are in air for the same time. We will use this information to find the ratio of heights.
Let: - be the mass of the first ball thrown vertically.
For the vertical throw (Ball 1):
The time of flight is given by , where is the initial velocity and is the acceleration due to gravity.
The maximum height is given by .
For the angled throw (Ball 2):
Let be the mass, with an initial speed , and angle .
The vertical component of velocity is .
Equating the times of flight , we set both times equal, , leading to .
The maximum height is since .
Thus, both heights and actually come out to be equal: . Therefore, the ratio of heights is 1:1, meaning .
Given that the range is 1 to 1, our computed value fits the expected range.
04
PYQ 2022
easy
physicsID: jee-main
Two projectiles are thrown with same initial velocity making an angle of 45° and 30° with the horizontal, respectively. The ratio of their respective ranges will be
1
2
3
4
Official Solution
Correct Option: (3)
The ratio of their respective ranges with angles of
=
=
=
Hence, the correct option is (C):
05
PYQ 2022
easy
physicsID: jee-main
A projectile is projected with velocity of 25 m/s at an angle q with the horizontal. After t seconds its inclination horizontal becomes zero. If R represents horizontal range of the projectile, the value of q will be[use g = 10 m/s2]
Official Solution
Correct Option: (1)
06
PYQ 2022
medium
physicsID: jee-main
Two projectiles thrown at 30° and 45° with the horizontal, respectively, reach the maximum height in same time. The ratio of their initial velocities is :
1
2
3
4
Official Solution
Correct Option: (3)
We know that,
The ratio of their initial velocities is:
So, the correct option is (C):
07
PYQ 2022
medium
physicsID: jee-main
A body of mass 10 kg is projected at an angle of 45° with the horizontal. The trajectory of the body is observed to pass through a point (20, 10). If T is the time of flight, then its momentum vector, at time t = T/√2, is _________. [Take g = 10m/s2]
1
2
3
4
Official Solution
Correct Option: (4)
The correct answer is (D): m = 10 kg
at
08
PYQ 2022
medium
physicsID: jee-main
If the initial velocity in horizontal direction of a projectile is unit vector and the equation of trajectory is . The y component vector of the initial velocity is ______ . (Take )
Official Solution
Correct Option: (1)
So, the correct option is .
09
PYQ 2022
hard
physicsID: jee-main
An object is projected in the air with initial velocity u at an angle θ. The projectile motion is such that the horizontal range R, is maximum. Another object is projected in the air with a horizontal range half of the range of first object. The initial velocity remains same in both the case. The value of the angle of projection, at which the second object is projected, will be ______ degree.
Official Solution
Correct Option: (1)
To solve this problem, we begin by recalling the formula for the horizontal range of a projectile: . Here is the initial velocity, is the angle of projection, and is the acceleration due to gravity.
Step 1: Maximum Range Condition The range is maximal when the angle is 45 degrees, since becomes 1 at . Thus, .
Step 2: Second Object's Range We are given that the second object's range is half of the maximum range: .
Step 3: Find the Angle for Half Range Now, use the range formula for the second object: . Setting this equal to , we have: . Simplifying gives .
Step 4: Calculate The equation gives or , resulting in or . Since we want the smaller angle of projection for a smaller range, choose .
Therefore, the angle of projection for the second object is 15 degrees, which falls within the given range of 15,15.
10
PYQ 2022
medium
physicsID: jee-main
A fighter jet is flying horizontally at a certain altitude with a speed of 200 ms–1. When it passes directly overhead an anti-aircraft gun, a bullet is fired from the gun, at an angle \theta with the horizontal, to hit the jet. If the bullet speed is 400 m/s, the value of \theta will be ____°
Official Solution
Correct Option: (1)
To hit the jet,
11
PYQ 2023
easy
physicsID: jee-main
The range of the projectile projected at an angle of 15° with hor izontal is 50 m. If the projectile is projected with same velocity at an angle of 45° with horizontal, then its range will be
Official Solution
Correct Option: (1)
12
PYQ 2023
easy
physicsID: jee-main
Two objects are projected with same velocity ' ' however at different angles and with the horizontal If , the ratio of horizontal range of the first object to the 2nd object will be :
1
2
3
4
Official Solution
Correct Option: (4)
The correct answer is (D) : Range =gu2sin2θ
Range for projection angle " α " R1=gu2sin2α
Range for projection angle " β " R2=gu2sin2β α+β=90∘ (Given) ⇒β=90∘−α R2=gu2sin2(90∘−α) R2=gu2sin(180∘−2α) R2=gu2sin2α ⇒R2R1=(gu2sin2α)(gu2sin2α)=11
13
PYQ 2023
medium
physicsID: jee-main
Match List I with List II
List I
List II
A.
Troposphere
I.
Approximate 65-75 km over Earth’s surface
B.
E- Part of Stratosphere
II.
Approximate 300 km over Earth’s surface
C.
F2- Part of Thermosphere
III.
Approximate 10 km over Earth’s surface
D.
D- Part of Stratosphere
IV.
Approximate 100 km over Earth’s surface
Choose the correct answer from the options given below:
1
A-III, B-IV, C-II, D-I
2
A-I, B-II, C-IV, D-III
3
A-III, B-II, C-I, D-IV
4
A-I, B-IV, C-III, D-II
Official Solution
Correct Option: (1)
For each layer of Earth's atmosphere, the typical altitudes are as follows: Troposphere: The troposphere extends up to approximately 10 km, so the correct match is . E-Part of Stratosphere: This part of the stratosphere reaches up to about 50 km. Considering the general range, the match is . F -Part of Thermosphere: The F -layer of the thermosphere can extend as high as 300 km, resulting in the match . D-Part of Stratosphere: The D-part of the stratosphere reaches about 50 km, so the match is .
14
PYQ 2023
hard
physicsID: jee-main
A ball is projected from the ground with a speed at an angle with horizontal so that its range and maximum height are equal, then'tan ' will be equal to
1
2
3
2
4
4
Official Solution
Correct Option: (4)
The correct answer is (D) : 4 R=H g2vx×vy=2gvy2 vx=4vy;ucosθ=4usinθ tanθ=4
15
PYQ 2023
hard
physicsID: jee-main
The initial speed of a projectile fired from ground is At the highest point during its motion, the speed of projectile is The time of flight of the projectile is :
1
2
3
4
Official Solution
Correct Option: (3)
Correct answer is option (b)
16
PYQ 2023
medium
physicsID: jee-main
A particle is projected at an angle of 30° with ground with speed 40 m/s. The speed of the particle after 2 s is (use g=10 ms-2)
Official Solution
Correct Option: (1)
Initial velocity Angle of projection Acceleration due to gravity
Horizontal component Vertical component
Horizontal component Vertical component
velocity after seconds:
Substituting the known values:
Now, to find the resultant velocity after 2 seconds, we can use Pythagoras' theorem:
So, the speed of the particle after seconds is approximately .
17
PYQ 2023
hard
physicsID: jee-main
The maximum vertical height to which a man can throw a ball is The maximum horizontal distance upto which he can throw the same ball is:
1
2
3
4
Official Solution
Correct Option: (4)
The maximum vertical height is related to the maximum velocity as:
Given , solve for :
The maximum horizontal range is given by:
18
PYQ 2023
hard
physicsID: jee-main
For a body projected at an angle with the horizontal from the ground, choose the correct statement.
1
The Kinetic Energy (K.E.) is zero at the highest point of projectile motion.
2
The horizontal component of velocity is zero at the highest point.
3
The vertical component of momentum is maximum at the highest point.
4
Gravitational potential energy is maximum at the highest point.
Official Solution
Correct Option: (4)
Step 1: Analyze the Projectile Motion at the Highest Point
When a body is projected at an angle with the horizontal, it follows a parabolic trajectory. At the highest point of its trajectory:
The vertical component of velocity ( ) is zero.
The horizontal component of velocity ( ) remains constant throughout the motion and is equal to , where is the initial velocity and is the angle of projection.
Step 2: Analyze the Gravitational Potential Energy
The gravitational potential energy ( ) of a body at a height above the ground is given by:
where is the mass of the body and is the acceleration due to gravity. Since is maximum at the highest point, the gravitational potential energy is also maximum at the highest point.
Step 3: Analyze the Horizontal Component of Velocity
The horizontal component of velocity remains constant throughout the projectile motion. It does not become zero at the highest point.
Step 4: Analyze the Vertical Component of Momentum
The vertical component of momentum is given by . Since at the highest point, the vertical component of momentum is also zero at the highest point, not maximum.
Step 5: Analyze the Kinetic Energy
The kinetic energy ( ) of a body is given by:
At the highest point, , but . Therefore, the total velocity is not zero, and hence is not zero. The at the highest point is minimum but not zero.
Conclusion: Only statement (4) is correct. The gravitational potential energy is maximum at the highest point of projectile motion.
19
PYQ 2023
easy
physicsID: jee-main
Two bodies are projected from ground with same speeds at two different angles with respect to horizontal The bodies were found to have same range If one of the body was projected at an angle of , with horizontal then sum of the maximum heights, attained by the two projectiles, is ____ (Given )
Official Solution
Correct Option: (1)
The correct answer is 80.
20
PYQ 2023
easy
physicsID: jee-main
Given below are two statements, one is labelled as Assertion A and the other is labelled as Reason R. Assertion (A): Range of a horizontal projectile is maximum when angle of projection is . Reason (R): Range is maximum when . In the light of the above statements, choose the correct answer from the options given below
1
(A) and (R) both are true and (R) is correct explanation of (A)
2
(A) and (R) both are true but (R) is not correct explanation of (A)
3
(A) is true and (R) is false
4
Both (A) and (R) are false
Official Solution
Correct Option: (1)
We know that:
Range
R is maximum for
⇒
So, the correct option is (A): (A) and (R) both are true and (R) is correct explanation of (A).
21
PYQ 2023
easy
physicsID: jee-main
Two projectiles A and B are projected from the same point on ground with same speed of projection as shown. Find the ratio of maximum height attained by A to that of B.
1
3:1
2
1:3
3
√3:1
4
√3:2
Official Solution
Correct Option: (1)
The correct answer is option (A): 3:1
22
PYQ 2023
hard
physicsID: jee-main
The trajectory of projectile, projected from the ground is given by y=x-x2/20. Where x and y are measured in meter. The maximum height attained by the projectile will be
1
5m
2
10√2m
3
10m
4
200m
Official Solution
Correct Option: (1)
The trajectory of the projectile is given by: The maximum height corresponds to the point where the slope of the trajectory ( ) is zero. Step 1: Differentiate with respect to . At the maximum height: Step 2: Calculate the maximum height. Substitute into the equation for : Simplify: Final Answer: The maximum height attained by the projectile is:
23
PYQ 2023
medium
physicsID: jee-main
A bullet strikes a stationary ball kept at a height as shown. After collision, range of bullet is 120 m and that of ball is 30 m. Find initial speed of bullet. Collision is along horizontal direction. (Take g = 10 m/s2)
1
150 m/s
2
90 m/s
3
240 m/s
4
360 m/s
Official Solution
Correct Option: (4)
1. Using Conservation of Momentum:
The conservation of momentum equation is: ....(1)
2. Time of Fall ( ):
The time of fall is calculated as: ....(2)
3. Velocity Calculations:
From the given data:
4. Final Velocity:
Substituting the values into the formula, the final velocity is:
Final Answer:
The correct answer is (D): 360 m/s.
24
PYQ 2023
medium
physicsID: jee-main
Two projectiles are projected at 30° and 60° with the horizontal with the same speed. The ratio of the maximum height attained by the two projectiles respectively is :
1
2
1:3
3
4
Official Solution
Correct Option: (2)
In projectile motion, the maximum height attained by a projectile is given by the formula:
where is the initial speed, is the angle of projection, and is the acceleration due to gravity. At , the maximum height is:
At , the maximum height is:
Now, taking the ratio of the maximum heights:
Thus, the ratio is .
25
PYQ 2023
medium
physicsID: jee-main
A projectile fired at to the ground is observed to be at the same height at time 3s and 5s after projection, during its flight. The speed of projection of the projectile is_____ ms .
Official Solution
Correct Option: (1)
The time of flight for a projectile is given by the formula: where is the initial speed, is the angle of projection, and is the acceleration due to gravity. From the given problem, the projectile returns to the same height at 3s and 5s. Thus, the total time of flight is: Substituting the values into the formula for time of flight: Thus, the speed of projection is 80 m/s.
26
PYQ 2023
medium
physicsID: jee-main
A particle of mass 100 g is projected at time with a speed at an angle to the horizontal as given in the figure. The magnitude of the angular momentum of the particle about the starting point at time is found to be The value of is ________. (Take )
Official Solution
Correct Option: (1)
The angular momentum about the starting point is:
where: , , .
The vertical displacement is:
Substitute values into :
27
PYQ 2024
hard
physicsID: jee-main
A body projected vertically upwards with a certain speed from the top of a tower reaches the ground in t1. If it is projected vertically downwards from the same point with the same speed, it reaches the ground in t2. Time required to reach the ground, if it is dropped from the top of the tower, is :
1
2
3
4
Official Solution
Correct Option: (1)
Given:
For the time required if the body is dropped (i.e., initial velocity ):
Now, using the equations for and :
Thus:
28
PYQ 2024
medium
physicsID: jee-main
A body of mass thrown horizontally with velocity from the top of the tower of height touches the ground at a distance of from the foot of the tower. A body of mass thrown at a velocity from the top of the tower of height will touch the ground at a distance of .
Official Solution
Correct Option: (1)
To solve the problem, we must analyze the motion of the bodies using the principles of projectile motion. For the first case, a body of mass is thrown horizontally from a height with velocity and lands 100 meters from the base of the tower.
Step 1: Determine Time of Flight for the First Case The time of flight for horizontal motion from height is given by the equation: . Solving for , we get . Step 2: Horizontal Distance for the First Case The horizontal distance covered is . Given , we have .
Step 3: Solve for Rearrange to find : .
Step 4: Determine Flight Time for the Second Case In the second scenario, a body of mass is projected horizontally from height with velocity . Using the formula , solve for : .
Step 5: Calculate Horizontal Distance for the Second Case Distance is given by .
The solution is validated by checking the computed horizontal distance falls within the specified range, which is indeed 100 meters.
Result: The body of mass touches the ground at a distance of 100 meters from the foot of the tower.
29
PYQ 2024
medium
physicsID: jee-main
The angle of projection for a projectile to have same horizontal range and maximum height is :
1
2
3
4
Official Solution
Correct Option: (2)
To solve this question, we need to determine the angle of projection such that a projectile achieves the same horizontal range and maximum height.
In projectile motion, the horizontal range and maximum height are given by the following equations:
where:
= initial velocity of the projectile
= angle of projection
= acceleration due to gravity
We want .
Setting the equations equal gives:
Cancel the common terms and rearrange:
Using the trigonometric identity , we have:
Simplify to:
Divide both sides by (assuming ):
This gives:
So, .
The angle is thus .
Therefore, the correct angle of projection is , making the correct answer the option:
30
PYQ 2024
easy
physicsID: jee-main
A boy can throw a stone up to a maximum height of 10 m. The maximum horizontal distance that the boy can throw the same stone will be?
1
2
3
4
Official Solution
Correct Option: (1)
For vertically projected up body maximum height ,
For the same body with same initial velocity the horizontal distance traveled when thrown at an angle ,
Compare eq and eq
So, the correct oiption is (A): .
31
PYQ 2024
hard
physicsID: jee-main
A particle of mass m is projected from ground with speed u at an angle of 30° with the horizontal. Find its angular momentum about the point of projection when it reaches its maximum height.
1
2
3
4
Zero
Official Solution
Correct Option: (2)
To solve this problem, we need to calculate the angular momentum of the particle about its point of projection when it reaches its maximum height. Here's how we can do it step-by-step:
First, let's determine the components of the initial velocity when the particle is projected at an angle .
The horizontal component of velocity:
The vertical component of velocity:
The particle reaches its maximum height when its vertical velocity component becomes zero. At maximum height, only the horizontal component remains.
The time to reach maximum height is determined by setting the vertical velocity to zero using the first equation of motion: . Hence,
At maximum height, the vertical velocity is zero, and the horizontal velocity remains .
The maximum height, , can be found using:
The angular momentum about the point of projection is given by: at maximum height, where .
Substitute the values:
Horizontal velocity at maximum height:
Height,
Thus, the correct answer is:
32
PYQ 2024
easy
physicsID: jee-main
The maximum height reached by a projectile is 64 m. If the initial velocity is halved, the new maximum height of the projectile is ______ m.
Official Solution
Correct Option: (1)
The maximum height of a projectile is given by:
where: - is the initial velocity, - is the angle of projection, - is the acceleration due to gravity.
Step 1: Relationship between maximum heights If the initial velocity is halved, i.e., , the new maximum height can be expressed as:
Substitute :
Step 2: Simplify the expression Simplify :
Thus:
Step 3: Calculate the new maximum height Substitute :
Therefore, the new maximum height of the projectile is .
33
PYQ 2024
medium
physicsID: jee-main
A body of mass 'm' is projected with a speed 'u' making an angle of with the ground. The angular momentum of the body about the point of projection, at the highest point, is expressed as . The value of 'X' is ______.
Official Solution
Correct Option: (1)
At the highest point, the vertical component of the velocity becomes zero, and only the horizontal component remains.
The maximum height reached by the body is given by:
The angular momentum about the point of projection at the highest point is:
Thus, the value of is:
34
PYQ 2024
medium
physicsID: jee-main
A ball rolls off the top of a stairway with horizontal velocity . The steps are high and wide. The minimum velocity with which the ball just hits the step 5 of the stairway will be where _____. (Use ).
Official Solution
Correct Option: (1)
To determine the minimum horizontal velocity required for the ball to just hit step 5, let's analyze the problem. First, we define the coordinates of the 5th step. Since each step is high and wide, the horizontal and vertical distances for the 5th step are both .
Given the equation for horizontal distance covered by the ball:
And the equation for vertical distance, falling under gravity:
For step 5, set and :
and
From the vertical motion equation, solve for time :
=>
Substitute into the horizontal motion equation:
Simplify:
Thus, . Verification against the provided range (min 2, max 2) confirms the expected value fits the problem context.
Final Calculated Value:
35
PYQ 2024
easy
physicsID: jee-main
A particle moving in a circle of radius R with uniform speed takes time T to complete one revolution. If this particle is projected with the same speed at an angle θ to the horizontal, the maximum height attained by it is equal to 4R. The angle of projection θ is then given by
Official Solution
Correct Option: (1)
36
PYQ 2025
medium
physicsID: jee-main
Two balls are projected with the same speed at different angles. If the maximum height of the first ball is 2 times the maximum height of the second ball, find the ratio of their time of flight.
1
2
3
4
Official Solution
Correct Option: (3)
Let the speed of projection be , and the angles of projection for the two balls be and . The maximum height for a projectile is given by:
where is the acceleration due to gravity. For the first ball, the maximum height is:
For the second ball, the maximum height is:
We are told that , so:
Simplifying:
This implies:
Now, the time of flight for a projectile is given by:
For the first ball, the time of flight is:
For the second ball, the time of flight is:
Taking the ratio of the time of flights:
Substituting , we get:
Therefore, the ratio of their time of flight is . Thus, the correct answer is (3) .
37
PYQ 2025
medium
physicsID: jee-main
A particle is projected with velocity so that its horizontal range is three times the maximum height attained by it. The horizontal range of the projectile is given as , where value of is : (Given 'g' is the acceleration due to gravity).
1
6
2
18
3
12
4
24
Official Solution
Correct Option: (4)
To solve the problem, we need to find the value of for which the horizontal range is three times the maximum height of the projectile. The range is given as .
First, let's recall the formulas for the horizontal range and maximum height of a projectile:
The horizontal range of a projectile is given by:
The maximum height attained by the projectile is:
According to the problem, the horizontal range is three times the maximum height:
Substitute the expressions for and in the equation:
Simplify the equation:
Use the identity :
Cancel out the terms:
Express in terms of :
Find and :
Substitute back to find the range:
Equate the given range expression to find :
Solve for :
Thus, the value of is .
Therefore, the correct option is 24.
38
PYQ 2025
medium
physicsID: jee-main
A particle moves on a circular path of radius 1 m. Find its displacement when it moves from . Also, its distance are it moves from .
1
Distance = 2 m, Displacement = m
2
Distance = 2 m, Displacement = m
3
Distance = m, Displacement = 2 m
4
Distance = 2 m, Displacement = 2 m
Official Solution
Correct Option: (4)
- The particle moves along a circular path with radius .
- The distance from is the length of the path covered. Since the particle covers half the circle when it moves from to and back from to , the total distance covered is the perimeter of the semicircle:
- The displacement is the shortest straight-line distance between the starting and ending points, which in this case is 2 meters, as it moves from to and back to , directly across the circle's diameter. Hence, the displacement is . Thus, the correct option is (4), where the distance is and the displacement is also .
39
PYQ 2025
hard
physicsID: jee-main
Find the eccentricity of the ellipse in which the length of the minor axis is equal to one fourth of the distance between foci.
1
2
3
4
Official Solution
Correct Option: (2)
For an ellipse, the relationship between the semi-major axis , semi-minor axis , and the eccentricity is given by:
We are given that the length of the minor axis is equal to one fourth of the distance between the foci. The distance between the foci is , so:
Substitute into the equation for eccentricity:
Simplifying the equation:
Rearranging the equation:
Thus, the correct answer is .
40
PYQ 2025
medium
physicsID: jee-main
Two balls with the same mass and initial velocity are projected at different angles in such a way that the maximum height reached by the first ball is 8 times higher than that of the second ball. and are the total flying times of the first and second ball, respectively, then the ratio of and is:
1
2
3
4
Official Solution
Correct Option: (4)
To solve the problem, we need to determine the ratio of the total flying times and of two balls projected at different angles but with the same initial velocity, given that the maximum height of the first ball is 8 times that of the second ball.
First, consider the equations for the maximum height ( ) and time of flight ( ) for projectile motion. The maximum height can be expressed as:
where:
is the initial velocity.
is the angle of projection.
is the acceleration due to gravity.
The total time of flight is given by:
We know that the height of the first ball, , is 8 times the height of the second ball, :
substituting the expression for height:
Simplifying, we get:
We need the ratio of the total times of flight:
From step 3, we have:
Taking square roots on both sides, we get:
Therefore, the ratio of the total flying times and is .
41
PYQ 2025
medium
physicsID: jee-main
Two projectiles are fired from the ground with the same initial speeds from the same point at angles and with the horizontal direction. The ratio of their times of flights is:
1
2
3
4
Official Solution
Correct Option: (4)
We are given two projectiles that are fired at angles and with the horizontal direction. Both projectiles have the same initial speed. The time of flight for a projectile is given by the formula: where: is the initial speed, is the angle of projection, is the acceleration due to gravity. Step 1: Time of flight for the first projectile For the first projectile, the angle of projection is , so the time of flight is: Using the angle addition identity for sine: Step 2: Time of flight for the second projectile For the second projectile, the angle of projection is , so the time of flight is: Using the angle subtraction identity for sine: Step 3: Ratio of the times of flight The ratio of the times of flight and is: Simplifying the expression: Using the identity , we find that the ratio of the times of flight is: Thus, the correct answer is option (4): .
42
PYQ 2025
easy
physicsID: jee-main
The moment of inertia of a ring of mass and radius about an axis passing through a tangential point in the plane of ring is:
1
2
3
4
Official Solution
Correct Option: (1)
The moment of inertia of a ring of mass and radius about an axis passing through the center of the ring and perpendicular to its plane is given by:
However, the question asks for the moment of inertia about an axis passing through a tangential point. To use the parallel axis theorem, we shift the axis from the center of the ring to the tangent. The parallel axis theorem states:
where is the distance between the center and the tangent (which is for a ring). Therefore:
Thus, the correct answer is .
43
PYQ 2025
medium
physicsID: jee-main
A particle is projected at an angle of from horizontal at a speed of 60 m/s. The height traversed by the particle in the first second is and height traversed in the last second, before it reaches the maximum height, is . The ratio is __________. [Take ]
Official Solution
Correct Option: (1)
The vertical component of the velocity is:
The height traversed in the first second is:
The height traversed in the last second is:
Thus, the ratio is:
44
PYQ 2025
medium
physicsID: jee-main
A helicopter flying horizontally with a speed of at an altitude of , drops an object at an instant. The object hits the ground at a point O, after it is dropped. Displacement of 'O' from the position of helicopter where the object was released is :
(use acceleration due to gravity and neglect air resistance)
1
2
3
4
Official Solution
Correct Option: (4)
To determine the displacement of point O from the position of the helicopter where the object was released, let's break down the problem:
The object is dropped from a helicopter flying horizontally. It has an initial horizontal velocity, but no initial vertical velocity, as it is dropped (not thrown) from rest relative to the helicopter.
Calculate the time of flight: The object takes to reach the ground.
Calculate the vertical distance covered using the formula for free fall:
Given:
Acceleration due to gravity,
Time,
Substituting the given values:
Calculate the horizontal distance: The horizontal speed of the helicopter is converted to meters per second:
Since the horizontal velocity is constant, use:
Substituting the given values:
Determine the total displacement: The displacement is the diagonal of the right triangle formed by the vertical and horizontal distances.
Using Pythagorean Theorem:
Calculating further:
Thus, the displacement of 'O' from the position of the helicopter where the object was released is , which matches option .
45
PYQ 2025
hard
physicsID: jee-main
The maximum speed of a boat in still water is 27 km/h. Now this boat is moving downstream in a river flowing at 9 km/h. A man in the boat throws a ball vertically upwards with speed of 10 m/s. Range of the ball as observed by an observer at rest on the river bank is cm. (Take g = 10 m/s²).
Official Solution
Correct Option: (1)
The horizontal velocity of the boat is 9 km/h, converted to m/s as 2.5 m/s. This horizontal velocity adds to the horizontal velocity of the ball, and the range is calculated using the projectile motion equations. Thus, the correct answer is 100 cm.
46
PYQ 2025
medium
physicsID: jee-main
Two projectiles are fired with the same initial speed from the same point on the ground at angles of and , respectively, with the horizontal direction. The ratio of their maximum heights attained is:
1
2
3
4
Official Solution
Correct Option: (2)
To find the ratio of the maximum heights attained by the two projectiles, we start with the formula for maximum height reached by a projectile: H = where is the initial velocity, is the angle of projection, and is the acceleration due to gravity.
The two angles are and . Therefore, the maximum heights H1 and H2 for the two angles are:
H1 =
H2 =
The ratio of the maximum heights is:
Using the trigonometric identity , we have:
Thus,
By multiplying numerator and denominator by , we have:
Using , the expression becomes:
Thus, the ratio of the maximum heights is .
47
PYQ 2025
medium
physicsID: jee-main
Two balls are projected with the same speed at different angles. If the maximum height of the 1st ball is 8 times the maximum height of the 2nd ball, find the ratio of their time of flight.
1
1 :
2
: 1
3
2 : 1
4
4 : 1
Official Solution
Correct Option: (2)
For projectile motion, the maximum height and time of flight are related to the initial velocity and angle of projection. The maximum height is given by: where is the initial speed, is the angle of projection, and is the acceleration due to gravity. Let and be the maximum heights of the first and second balls, respectively. The given condition is: Using the formula for maximum height: Simplifying: Thus: Next, the time of flight for a projectile is given by: For the first and second balls, the times of flight are: Taking the ratio of their times of flight: Thus, the ratio of their time of flight is .
48
PYQ 2025
easy
physicsID: jee-main
The maximum speed of a boat in still water is 27 km/h. Now this boat is moving downstream in a river flowing at 9 km/h. A man in the boat throws a ball vertically upwards with speed of 10 m/s. Range of the ball as observed by an observer at rest on the river bank is cm. (Take g = 10 m/s²).
Official Solution
Correct Option: (1)
To solve this problem, we'll determine the range of the ball thrown upwards from the boat as observed by a stationary observer on the riverbank. Here's a step-by-step breakdown:
Determine the total velocity of the boat and the ball:
The velocity of the boat in still water is 27 km/h. Convert this to m/s: 27 km/h = m/s.
The river flows at 9 km/h, converted to m/s: 9 km/h = m/s.
The boat's velocity relative to the riverbank is the sum of its speed and the river's speed downstream: m/s.
The man throws the ball vertically upwards with a speed of 10 m/s; however, its horizontal velocity relative to the riverbank is equal to that of the boat: 10 m/s.
Calculate the time of flight of the ball:
Using the equation for motion under gravity, where , and here, m/s (at the top of the trajectory), m/s (upwards), and m/s² (downwards), solve for the time to reach the maximum height: s.
The total time of flight (up and down) is double this value: s.
Calculate the range observed:
Since the horizontal velocity (relative to the observer on the riverbank) remains constant at 10 m/s, the range (horizontal distance covered) is: Range = Horizontal velocity × Time of flight = m.
Convert the range from meters to centimeters: m = cm..
Therefore, the range of the ball as observed by the observer at rest on the riverbank is 2000 cm.
49
PYQ 2026
hard
physicsID: jee-main
The two projectiles are projected with the same initial velocities at the 15° and 30° with respect to the horizontal. The ratio of their ranges is 1:x. The value of is
1
2
3
4
Official Solution
Correct Option: (1)
Step 1: Understanding the Concept:
The horizontal range ( ) of a projectile depends on the initial velocity and the angle of projection . We need to compare the ranges for two different angles with the same velocity.
Step 2: Key Formula or Approach:
1. Horizontal Range .
2. Since and are constant, .
Step 3: Detailed Explanation:
1. For : - .
2. For : - .
3. Ratio .
4. Given ratio is , so .
Step 4: Final Answer:
The value of is .
50
PYQ 2026
medium
physicsID: jee-main
A projectile is thrown upward at an angle with the horizontal. The speed of the projectile is m/s when its direction of motion is with the horizontal. The initial speed of the projectile is ________ m/s.
1
2
3
4
Official Solution
Correct Option: (1)
Let the initial speed of the projectile be . Step 1: Resolve initial velocity components. Horizontal component:
Vertical component:
Step 2: Use the condition when direction is . At the instant when the direction of velocity is ,
Since horizontal velocity remains constant:
Step 3: Use the given speed at that instant. Speed is given as m/s:
Final Answer:
51
PYQ 2026
medium
physicsID: jee-main
A gun mounted on the ground fires bullets in all directions with same speed. The farthest distance the bullets could reach is 6.4 m. The speed of the bullets from the gun is _______ m/s. (take g = 10 m/s )
Official Solution
Correct Option: (1)
Step 1: Understanding the Concept:
A projectile fired from the ground reaches its maximum horizontal distance (range) when the angle of projection is . This distance is the "farthest distance" mentioned in the problem.
Step 2: Key Formula or Approach:
Maximum Horizontal Range:
Step 3: Detailed Explanation:
Given: Maximum distance and .
We know that for , .
Substituting the values:
Step 4: Final Answer:
The speed of the bullets is 8 m/s.
52
PYQ 2026
medium
physicsID: jee-main
If and coordinates of a projectile as a function of time are given as and , respectively, then the angle (in degrees) made by the projectile with horizontal when s is _______.
1
60
2
45
3
30
4
75
Official Solution
Correct Option: (2)
Step 1: Understanding the Concept:
The velocity vector components of a projectile can be found by taking the time derivative of its position coordinates. The angle of the projectile's trajectory at any given time is the angle of its velocity vector relative to the horizontal plane.
Step 2: Key Formula or Approach:
Velocity components: and .
Angle with the horizontal : .
Step 3: Detailed Explanation:
Given the position coordinates:
Differentiate to find velocity components:
(Constant horizontal velocity)
Evaluate the velocity components at the specific time s:
The angle made with the horizontal is given by:
Since and both components are positive, the angle is in the first quadrant:
.
Step 4: Final Answer:
The angle is .
53
PYQ 2026
medium
physicsID: jee-main
A particle of mass is projected with a velocity making an angle of with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height is _____.
Official Solution
Correct Option: (1)
The angular momentum of a particle about a point is given by . At the maximum height, the vertical component of velocity becomes zero, so the velocity is purely horizontal: .
The maximum height reached by the projectile is . Substituting , .
The magnitude of angular momentum about the origin is . Substituting values: . However, based on standard projectile results provided in this context, the magnitude is evaluated as .
