A ball is dropped from a height of `320 m` above the ground. After every collision, the speed of ball decreases by `50%`. Taking dropping point as origin, downward direction positive and collision time negligible, sketch the v-t, s-t and a-t graphs. Also calculate the total distance traveled by the ball and the total time of journey.

Explanation

O to A: `s=ut+1/2g t^(2)`
`320=0+1/2xx10xxt_(0)^(2)impliest_(0)=8s`
`v=u+g t`
`v_(0)=0+10 t_(0)=80 m//s`
Speed of the ball after first collision `=1/2 (80)=40 m//s=v_(1)`
A to B: `0=v_(1)^(2)-2gh_(1)implies0=(40)^(2)-20h_(1)impliesh_(1)=80 m`
`0=v_(1)-g t_(1)implies0=40-10 t_(1)impliest_(1)=4 s`
Speed of the ball after second collision`=1/2xx40 .....=20 m//s=v_(2)`
A to C: `0=v_(2)^(2)-2gh_(2)implies0=(20)^(2)-2xx10xxh_(2)impliesh_(2)=20 m`
`0=v_(2)-g t_(2)implies0=20-10t_(2)impliest_(2)=2s`
and so on.
After every collision, the speed becomes half, the maximum height becomes `1//4^(th)` and the time to reach at the highest point becomes half.