A ball is dropped from a height of 90 m on a floor. At each collision with the floor, the ball loses one tenth of its speed. Plot the speed-time graph of its motion between `t =0` to 12 s.

Explanation

Here initial speed `v_(0)=0 h = 90 m`
Lost speed in each collision, `=1/10 xx ` Speed acquired while striking ground,
If the velocity of ball while striking ground is `v_(1),` then
`v _(1) ^(2) - v _(0) ^(2) = 2gh`
`v _(1) ^(2-) -0 = 2 xx 9.8 xx 90`
`v _(1) ^(2) sqrt (2 xx 9.8 xx 90)`
` v _(1) = 42 ms ^(-1)`
and time taken to reach the ground is `t_(1), `then
From `d = v_(0) t + 1/2 g t ^(2) `
`h = 1/2 g t _(1) ^(2)`
`t _(1) = sqrt ((2g)/(g))`
` = sqrt ((2 xx 90)/(9.8))`
`t _(1) = 4.28 s`
If the velocity when returning back after striking the ground is `v _(2),` then
`v _(2) = v _(1) - (10 % of v _(1))`
`= 42 -4.2`
`= 37.8 ms ^(-1)`
Height achieved by ball after striking ground is `h_(1),` then
`h _(1) = (v _(2) ^(2))/( 2g) = (37.8 xx 37.8)/(2 xx 9.8) = 72.9m`
Now when ball again strikes the ground, velocity is `V_(3),` then its displacemtn will be O.
So, `v_(3) ^(2) - v _(2) ^(2) = 2g xx 0 [ because h =0]`
`therefore v _(3) ^(2) = v _(2) ^(2)`
`v _(3) = v _(2) = 37.8 ms ^(-1)`
If time taken between two successive colisions is `t _(2),` then
`0= v _(2) t _(2) + 1/2 (-g) t_(2) ^(2)`
`0= 37.8 t _(2) - (9.8)/(2) t _(2) ^(2)`
`4.9 t _(2) ^(2) = 37.8 t _(2)`
`t _(2) = 7.714 s.`
If velocity after second collision is `v _(4),` then
`v _(4) = v _(3) - 10% of v _(3)`
`= 37.8 -3.78`
`v _(4) = 34.02 ms ^(-1)`
Now `t _(2) = 7.714s` is time of flight (2t)
`[because` Time taken to reach maximum height]
={Time taken to return back}
So that `t _(2) = 2t.`
`t. = (t_(2))/(2)`
` t . = ( 7.714)/(2) =3.857=3.86s`
Now `t_(1) + t _(2) =4.28 + 7.714=11.994 s=12s`

Thus, ball will return back in `t =0` to t =12 s after colliding ground.
Velocity of ball is `37.8 ms ^(-1)` when strikes first time on ground.