# Q 3.12 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.

At time t = 0 velocity is 0

Initial velocity , u = 0

Acceleration = 10 ms-2

Height, s = 90 m

$\\s=ut+\frac{1}{2}at^{2}\\ 90=0\times t+\frac{1}{2}\times 10t^{2}\\ t=3\sqrt{2}s\\ t=4.242s$

$\\v=u+at\\ v=10\times 3\sqrt{2}\\v=30\sqrt{2}ms^{-1}$

The above is the speed with which the ball will collide with the ground. After colliding the upwards velocity becomes

$\\u=30\sqrt{2}\times \frac{9}{10}\\ u=27\sqrt{2}ms^{-1}$

Acceleration a = -10ms-2

While the ball will again reach the ground its velocity would have the same magnitude $v=-27\sqrt{2}ms^{-1}$

Let the time between the successive collisions be t

$\\v=u+at\\ t=\frac{v-u}{a}\\ t=\frac{-27\sqrt{2}-27\sqrt{2}}{-10}\\ t=5.4\sqrt{2}s\\ t=7.635$

After the first collision, its speed will become 0 in time $2.7\sqrt{2}s$

Total time = 4.242 + 7.635 = 11.87s

After this much time, it will again bounce back with a velocity v given by

$\\v=27\sqrt{2}\times \frac{9}{10}\\ v=24.3\sqrt{2}ms^{-1}$

$v_{E}=24.3\sqrt{2}ms^{-1}$

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