Motion In A Plane

Step 1: The formula for the time period of a simple pendulum is given by: where is the length of the pendulum and is the acceleration due to gravity. Step 2: Gravity decreases with height. The formula for at a height above the earth’s surface is: where is the radius of the Earth. Step 3: The ratio of the time periods at the surface and at height is: Substitute the given values: and : Thus, the time period at the height is .

Step 1: Using the kinematic equation for vertical motion The equation of motion for vertical displacement: At maximum height , the final velocity is zero: Solving for : Now, considering the time to reach height , we use: Substituting : Rearranging: Step 2: Solving the quadratic equation Dividing throughout by : Using the quadratic formula : Step 3: Finding the ratio of the times Since the two possible values of are: The required ratio is: Step 4: Verifying the correct option Comparing with the given options, the correct answer is:

Step 1: Compute time of flight
Vertical velocity: Solving for time of flight : Step 2: Compute horizontal range
Horizontal velocity: Thus, the correct answer is m.

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Step 1: Time of Flight Formula

For , , solving gives:
Step 3: Conclusion

Thus, Option (3) is correct.

Step 1: Use toppling condition
The toppling condition occurs when the moment due to centrifugal force equals the moment due to weight: Step 2: Solve for
Thus, the correct answer is ms .

Step 1: Vector Addition:
First, we find the resultant vector . Given: Adding corresponding components:
Step 2: Magnitude of Resultant Vector:
The magnitude is given by .
Step 3: Calculating the Unit Vector:
The unit vector along is defined as . We can simplify this by factoring out from the numerator:
Step 4: Final Answer:
The unit vector is .

Step 1: Compare the given trajectory equation with the standard form.
The standard equation of a projectile's trajectory is: where is the launch angle and is the initial speed. The given equation is .

Step 2: Find the launch angle and initial speed.
By comparing the coefficients: . . Substitute and (so ):

Step 3: Calculate the time of flight.
The formula for the time of flight, , is: Substitute the values we found: .

Let the initial velocity of the projectile be . The angle of projection is .
Let the ball be at a certain height at times s and s.
The equation for the vertical position of a projectile is .
For a given height , we have , which can be rearranged into a quadratic equation in :
.
The roots of this equation are the times when the projectile is at height . We are given these times as s and s.
From the properties of quadratic equations, the sum of the roots is .
We know that the total time of flight of a projectile is .
Therefore, the total time of flight is s.
The horizontal motion of a projectile is uniform motion with velocity .
The horizontal distance between the two points is the distance traveled horizontally between s and s.
Horizontal distance .
We need to find . From the time of flight formula:
m/s.
Now, m/s.
Finally, the horizontal distance is m.

Step 1: Understanding the Concept:
This is a projectile motion problem. We are given the initial velocity and maximum height and need to find the horizontal range. We can use the formula for maximum height to find the angle of projection first, and then use that angle to find the range. We'll use .
Step 2: Key Formulae:
Maximum Height:
Horizontal Range:
Given: m/s, m.
Step 3: Detailed Explanation:
First, we find the angle of projection, , using the maximum height formula.
Since , we can write:
This implies that the angle of projection is .
Now, we can calculate the range using the range formula and .
Since :
Step 4: Final Answer:
The range of the projectile is 39.2 m. Therefore, option (B) is correct.

• Vertical motion: initial y = 55 m, m/s, m/s\textsuperscript{2} downward. Solve .
• This gives s (positive root).
• Horizontal displacement: m.
• Vertical displacement (change) = final y - initial y = m → represented as .
• Hence displacement vector = m → option (1).

Let , . For a projectile, , . Using kinematics, time of flight → solves to .