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.

Answers (1)

At time t = 0 velocity is 0

Initial velocity , u = 0

Acceleration = 10 ms-2

Height, s = 90 m

\s=ut+frac12at^2\ 90=0	imes t+frac12	imes 10t^2\ t=3sqrt2s\ t=4.242s

\v=u+at\ v=10	imes 3sqrt2\v=30sqrt2ms^-1

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

\u=30sqrt2	imes frac910\ u=27sqrt2ms^-1

Acceleration a = -10ms-2

While the ball will again reach the ground its velocity would have the same magnitude v=-27sqrt2ms^-1

Let the time between the successive collisions be t

\v=u+at\ t=fracv-ua\ t=frac-27sqrt2-27sqrt2-10\ t=5.4sqrt2s\ t=7.635

After the first collision, its speed will become 0 in time 2.7sqrt2s

Total time = 4.242 + 7.635 = 11.87s

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

\v=27sqrt2	imes frac910\ v=24.3sqrt2ms^-1

 

 v_E=24.3sqrt2ms^-1

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