$A$ 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 $12 \; s$.

(N/A) The ball is dropped from a height $s = 90 \; m$.
Initial velocity $u = 0$,acceleration $a = g = 9.8 \; m/s^2$.
Time $t$ taken to hit the ground: $s = ut + (1/2)at^2 \implies 90 = 0 + (1/2) \times 9.8 \times t^2 \implies t = \sqrt{18.367} \approx 4.29 \; s$.
Final velocity just before the first impact: $v = u + at = 0 + 9.8 \times 4.29 = 42.04 \; m/s$.
Rebound velocity $u_r = (9/10)v = 0.9 \times 42.04 = 37.84 \; m/s$.
Time to reach maximum height after first rebound: $v = u_r + at' \implies 0 = 37.84 - 9.8 \times t' \implies t' = 3.86 \; s$.
Total time for the ball to return to the floor: $t + 2t' = 4.29 + 2 \times 3.86 = 12.01 \; s$.
Velocity just before the second impact: $v_2 = 37.84 - 9.8 \times 3.86 = 0 \; m/s$ (at peak),then it accelerates back down to $37.84 \; m/s$ at the floor.
After the second collision,the new rebound velocity is $u_{r2} = (9/10) \times 37.84 = 34.06 \; m/s$.