Question

A ball is dropped from a height of 19.6 m above the ground. It rebounds from the ground and raises itself up to the same height. Take the starting point as the origin and vertically downward as the positive X-axis. Draw approximate plots of x versus t, v versus t and a versus t. Neglect the small interval during which the ball was in contact with the ground. 

Answer 1

From the first equation of motion for constant acceleration, plot the v-t graph. From the second equation of motion for constant acceleration, plot the x-t graph.

Since the acceleration of the ball during the contact is different from ‘g’, we have to treat the downward motion and the upward motion separately.  

For the downward motion: a = g = 9.8 m/s² ,

The ball reaches the ground when x = 19.6 m. This gives t = 2 s. After that it moves up, x decreases and at t = 4 s, x becomes zero, the ball reaching the initial point. 

During the next two seconds the ball goes upward, velocity is negative, magnitude decreasing and at t = 

At t = 2 s there is an abrupt change in velocity from 19.6 m/s to -19.6 m/s. In fact this change in velocity takes place over a small interval during which the ball remains in contact with the ground.

Acceleration: The acceleration is constant 9.8 m/s² s throughout the motion (except at t=2s).