A particle of mass m is released from a certain height h with zero initial velocity. It strikes the ground elastlcally (direction of its velocity is reversed but magnitude remains the same). Plot the graph between its kinetic energy and time till it returns to its initial position.

Explanation

To solve the problem of plotting the graph between kinetic energy and time for a particle of mass \( m \) released from a height \( h \) with zero initial velocity, we can follow these steps:

### Step 1: Understand the motion of the particle
The particle is released from rest, so its initial velocity \( u = 0 \). It falls under the influence of gravity until it strikes the ground. The time taken to fall to the ground can be determined using the equations of motion.

### Step 2: Calculate the time to reach the ground
Using the second equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
where:
- \( s = h \) (the height from which it is released),
- \( u = 0 \) (initial velocity),
- \( a = g \) (acceleration due to gravity),
- \( t = t_0 \) (time taken to reach the ground).

Substituting the values, we have:
\[ h = 0 \cdot t_0 + \frac{1}{2} g t_0^2 \implies h = \frac{1}{2} g t_0^2 \implies t_0 = \sqrt{\frac{2h}{g}} \]

### Step 3: Calculate the velocity just before impact
Using the first equation of motion:
\[ v = u + at \]
Substituting the values:
\[ v = 0 + g t_0 = g \sqrt{\frac{2h}{g}} = \sqrt{2gh} \]

### Step 4: Calculate the kinetic energy just before impact
The kinetic energy \( KE \) just before the particle hits the ground is given by:
\[ KE = \frac{1}{2} m v^2 = \frac{1}{2} m (2gh) = mgh \]

### Step 5: Analyze the motion after the impact
Since the collision is elastic, the particle will bounce back with the same speed \( v = \sqrt{2gh} \) but in the opposite direction. The time taken to return to the original height \( h \) will also be \( t_0 \).

### Step 6: Determine the kinetic energy during the motion
- From \( t = 0 \) to \( t = t_0 \): The kinetic energy increases from \( 0 \) to \( mgh \).
- From \( t = t_0 \) to \( t = 2t_0 \): The kinetic energy decreases from \( mgh \) back to \( 0 \).

### Step 7: Plot the graph
- The x-axis represents time \( t \).
- The y-axis represents kinetic energy \( KE \).
- The graph will be a parabola opening upwards, starting from \( (0, 0) \) to \( (t_0, mgh) \) and then back to \( (2t_0, 0) \).

### Summary of the graph:
- At \( t = 0 \), \( KE = 0 \).
- At \( t = t_0 \), \( KE = mgh \).
- At \( t = 2t_0 \), \( KE = 0 \).