A ball is dropped from a height of a height of 90 m on a floor. At each collsion 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

Taiking vertical downward motion of ball from a height ` 90 m`, we have
` u= 0, a =10 m//s^2 , S=90 m, t=? , v=?`
` t= sqrt ( 2S)/ a = sqrt( 2 xx 90)/910) = 3 sqr 2 s = 4. 24 s`
` v = sqrt 2 a S = sqrt (2 xx 10 xxx 90 ) = 30 sqrt 2 m//s`
Rebound velocity of ball, ` u' = 9/ (10) v = 9/ (10) xx 30 sqrt 2 = 27 sqrt2 m//s`
Time to reach the hight the highest point is , ` t' = u' /a = ( 27 sqrt 20 / (10) = 2.7 sqrt 2 = 3.81 s`
Total time ` = t+ t' = 4.24 + 3. 81 = 8 .05 s`
The ball will take further ` 3. 81 s` to fall back to floor, wher its velocity b efore striking the floor
` = 27 sqrt 2 m//s`.
Veloity of ball after striking the floor
` = 9/(10) xx 27 sqrt 2 = 24 .3 sqrt 3 m//s`
Total time elapsed before upward motion of ball
` = 8.05 + 3. 81 = 11 . 86 s`
Thus the speed-time graph of this of motion will be as shown in Fig. 2 (NCT).3.
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