Question

A particle of mass m is released from a certain height h with zero initial velocity. It strikes the ground elastically (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.

Answer 1

Initial velocity is zero then KE also becomes zero

now velocity of particle increases due to gravitation acceleration downward

$v=\sqrt{2gh}$ final velocity

then KE at ground is = $=\frac{1}{2}m{v}^2=mgh$

after collision with ground the velocity direction changes but magnitude remains the same as $v=\sqrt{2gh}$

so the KE also remains same

Now, the velocity is opposite to the gravitation acceleration so velocity decreases to zero after the same time

and KE also becomes at the highest point at height h

So, the graph of KE vs time is Shown in fig.