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.

Explanation

Since the aceleration of theball during the cobtact 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/s^2,
`x=ut+1/2 at^2=(4.9 m/s^2)t^2`.
the ball reaches the ground when x=19.6 m. this gives t=2s. After that it moves up, x decreases and at t=4 s,x becomes zero, the ball reaching the initial point.
We have t t=0,
, x=0
t=1, x=4.9 m
t=2s, x=19.6m
t=3s, x=4.9m
t=4s, x=0.
.
Velocity : During the first two seconds
`v=u+at=(9.8 m/s^2)t`
`at t=0, v=0`
`at t=1s, v=9.8 m/s`
at t=2s, v=19.6 m/s`
During the next two seconds the ball goes upwrd, velocity is negtive, magnitude decreasing and tat `t=4, s v=0` Thus,
`at t=2, v=-19.6 m/s`
at t=3s, v=-9.8 m/s
at t=4s, v=0

At t=2s there is an abrupt change in velocilty from 19.6 m,/s to -19.6 m/s In fasct this change in velocity takes plce over a small interval during which the ball remains in contact with the ground.
Acceleration: THe acceleration is constant `9.8 m/s^2` throughtout the motion (except t t=2s)`.