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Question
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.
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Solution 1
Ball is dropped from a height, s = 90 m
Initial velocity of the ball, u = 0
Acceleration, a = g = 9.8 m/s2
Final velocity of the ball = v
From second equation of motion, time (t) taken by the ball to hit the ground can be obtained as:
`s = ut + 1/2 at^2`
`90 = 0 + 1/2 xx 9.8t^2`
`t = sqrt(18.38) = 4.29 s`
From first equation of motion, final velocity is given as:
v = u + at
= 0 + 9.8 × 4.29 = 42.04 m/s
Rebound velocity of the ball, ur = `9/10v = 9/10 xx 42.04 = 37.84 "m/s"`
Time (t) taken by the ball to reach maximum height is obtained with the help of first equation of motion as:
v = ur + at′
0 = 37.84 + (– 9.8) t′
t' = (-37.84)/-9.8 = 3.86 s
Total time taken by the ball = t + t′ = 4.29 + 3.86 = 8.15 s
As the time of ascent is equal to the time of descent, the ball takes 3.86 s to strike back on the floor for the second time.
The velocity with which the ball rebounds from the floor = `9/10xx37.84 = 34.06 "m/s"`
Total time taken by the ball for second rebound = 8.15 + 3.86 = 12.01 s
The speed-time graph of the ball is represented in the given figure as:
Solution 2
u = 0, a = 10 ms-2, S = 90 m, t = ?, v = ?
Using `v^2 - u^2 = 2as, v^2 - (0)^2 = 2xx10xx90`
=> v = `30sqrt2` "m/s"
Again using `S = ut + 1/2at^2, 90 = 0 xx t + 1/2 xx 10t^2`
`=> t= sqrt18 s = 3sqrt2 s`
Rebound velocity = `9/10 xx 30sqrt2 ms^(-1) = 27sqrt2 ms^(-1)`
Time taken to reach highest point = `(27sqrt2)/10` s = `2.7sqrt2` s
Total time = `(3sqrt2 + 2.7sqrt2)s = 5.7sqrt2 s`
OA represents the vertically downward motion after the ball has been dropped from a height of 90m. The reaches the floor with a velocity of `30sqrt2 ms^(-1)` after having been in motion for `3sqrt2 s`. The verticle straight portion AB represent the loss of 1/10 th of speed, BC represents the vertically upward motion after first rebound. The ball reaches the highest point in `27sqrt2`s. The total time from the beginning is `3sqrt2+2.7sqrt2` i.e `5.7sqrt2` s.
C represent the highest point reached after first rebound. CD represent the vertically downward motion. D represents the situation when the ball again reaches the floor DE represent the loss of speed.
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