• Given a ball is thrown vertically down with a velocity of 10m/s from a height of 3m

• The coefficient of restitution is 0.7

   

To find

• The maximum height$(h)$ ball can reach after hitting the floor

   

Solution:

• Let the velocity of the ball after bouncing from the surface be$v_2$
• We know that$v_1^2-u^2=2gh$
• We get$v_1^2$=158.8

$v_1=12.61m/s$

• Given coefficient of restitution$ C=0.7$
• Let the velocity of the ball after bouncing from the surface be$v_2$

$ C={v_2 \over v_1}$=$\dfrac {\mathrm {relative \ velocity \ after\ impact}}{\mathrm {the \ relative \ velocity \ before \ the \ impact}}$

• We get$v_2=$0.7×12.61=8.82m/s
• Let the maximum height reached after hitting the floor be$h $ from the floor
• From$v^2-v_2^2=-2gh$

$v=0m/s$ (at maximum height)

$8.82×8.82=2×9.8×h$

We get $h=3.97m$

• So the maximum height (h) ball can reach after hitting the floor is$h=3.97m$