Projectile Motion

For a projectile projected at angle ;
Maximum height,
and range,
Ratio
Here,

For a projectile to cover a horizontal range , the formula is: The range will be maximum when . So,

Maximum height occurs when : Substitute in y: Correction: Answer should be (3) — mismatch in answer key noted.

Step 1: Given Step 2: Integrate to find Step 3: Apply initial condition: At , Final Result:

We are given that: For a projectile: Set : Using identity: ,

We are given the time of flight is 8 s, and the acceleration due to gravity is . The time taken to reach the ground is given by: The time taken to reach the point after 6 seconds is the remaining time for the upward motion, so: For vertical motion under gravity, we can use the following kinematic equation: where is the height, is the initial velocity, is the acceleration due to gravity, and is the time. We can calculate the position of the stone after 6 seconds. Using the equation of motion, the position after 6 seconds is:

For a vertically projected body, the velocity-time graph is typically a straight line that decreases as the body moves upward, eventually reaching zero velocity at the peak of the motion. The correct graph for such motion is option (2). Thus, the correct answer is option (2).

By comparing the given equations for vertical and horizontal motion of the projectile, we can calculate the initial velocity using the standard equations for projectile motion. The initial velocity is determined to be 5 m/s. Thus, the correct answer is option (3).

For a projectile launched at , the maximum range is given by .
The maximum height is given by at .
Thus, since .

Step 1: Calculate the maximum height of the first body (12 kg).
The body of mass 12 kg is projected vertically upwards with velocity . Using the formula for maximum height in vertical motion:
Step 2: Maximum height of the second body (6 kg).
The body of mass 6 kg is projected horizontally. Its vertical displacement is just the height of the tower, i.e., .
Step 3: Calculate the maximum height of the center of mass.
Using the center of mass formula: Substituting the values: Thus, the maximum height reached by the center of mass is approximately .

The vertical component of velocity is given by: Using the equation for vertical displacement: Substitute the values , , and : The height is 1 m above the ground.

We are given that a projectile can have the same range for two angles of projection with the same initial velocity. Let and be the times of flight for two different angles of projection. We are asked to find the relationship between the product of the two times of flight and the range . The time of flight for a projectile is given by the formula: where is the initial velocity, is the angle of projection, and is the acceleration due to gravity. The range of the projectile is given by: Since the initial velocity is the same for both cases, we can write: The product of the times of flight for two angles is: We can simplify this expression as: Using the identity and knowing that the range is directly proportional to , we can conclude that the product of the times of flight is directly proportional to . Thus, the correct answer is option (4).

Step 1: Define initial velocity and velocity at maximum height for a projectile.
Let the initial speed of projection be u, and the angle of projection with the horizontal be .
The initial velocity components are: Horizontal component (u\textsubscript{x}) = u cos Vertical component (u\textsubscript{y}) = u sin At the maximum height of a projectile's trajectory, the vertical component of velocity becomes zero (v\textsubscript{y} = 0). Only the horizontal component of velocity remains, which is constant throughout the projectile's flight (assuming no air resistance).
So, the speed at maximum height (v\textsubscript{max height}) = u\textsubscript{x} = u cos . Step 2: Use the given condition to set up an equation.
The problem states that the speed of projection (u) is twice its speed when it is at maximum height (u cos ).
So, Step 3: Solve the equation for the angle of projection ( ).
Divide both sides by u (assuming u 0):
To find , we take the inverse cosine: We know that cos(60°) = .
Therefore, °.
Step 4: Select the correct option.
The angle of projection is 60°, which matches option (4).

Step 1: Define average torque and angular momentum.
The average torque ( ) acting on a body is equal to the total change in angular momentum ( ) divided by the total time taken ( ).
Here, is the total time of flight. Angular momentum ( ) is defined as , where is the position vector from the origin, is the mass, and is the velocity. We choose the origin as the point of projection. Step 2: Calculate initial angular momentum ( ).
At the point of projection (initial time ), the position vector .
Therefore, the initial angular momentum . Step 3: Calculate the time of flight ( ).
For a projectile launched with initial velocity at an angle with the horizontal, the time of flight is given by: Step 4: Calculate final position and velocity at time of flight ( ).
At time , the projectile returns to the horizontal level (ground, ).
The horizontal distance traveled (range) is:
So, the final position vector is . (assuming projection is from origin, and horizontal is x-axis) The final velocity components are:

So, the final velocity vector is . Step 5: Calculate final angular momentum ( ).
Using the cross product properties ( and ): Step 6: Calculate the average torque.
Cancel common terms ( ): The magnitude of the average torque is . The final answer is .

The range of a projectile is given by: Given , , : Option (1) is correct. Options (2), (3), and (4) do not satisfy .

The time to reach point P in upward journey is . The total time to reach maximum height is . The ball takes equal time to come down from maximum height to point P. Hence, total time after point P is twice the time from to . So, time taken to pass through point P again is:

Given: Initial speed, , Angle, , Gravity, . Vertical component of velocity: Maximum height: Nearest option is 20 m.

The height of a projectile at time is: Given at , : The time of flight is: Option (4) is correct. Options (1), (2), and (3) do not match.