a ball is dropped from a height of 90m on a floor.at each collision with the floor ,the ball loses one tenth of ts speed.plot the speed-time graph of its motion between t=0 to 12s. please explain clearly.

Time taken by the ball to fall through a height of 90 m is obtained as follows: or Now From time t = 0 to t = 4.3 s, or In this duration speed increases linearly with time t from 0 to during the downward motion of the ball and this speed-time variation has been shown by straight line OA in Fig. 3.86. Fig. 3.86 (ii) At first collision with the floor, speed lost by ball . Thus, the ball rebounds with a speed of . For the further upward motion, the speed at any instant t is given by Now the speed decreases linearly with time and becomes zero after time Thus, the ball reaches the highest point again after time t = 4.3 + 3.9= 8.2s from the start. Straight line BC represents the speed-time graph for this upward motion. (iii) At highest point, speed of ball is zero. It again starts falling. At any instant t, its speed is given by Again the speed of the ball increases linearly with time t from 0 to 37.8 ms~1 (initial speed of the previous upward motion) in the next time-interval of 3.9 s. Total time taken from the start = 4.3 + 3.9 + 3.9 = 12.1 s. This part of motion has been shown by straight line CD. Here, we have assumed a negligible time of collision between the ball and the floor.