Question 7.SP.40: A ball rebounds vertically from a horizontal floor to a heig......
A ball rebounds vertically from a horizontal floor to a height of 20 m. On the next rebound, it reaches a height of 14 m. What is the coefficient of restitution between the ball and the floor?
Let subscript 1 refer to the ball. The initial and final speeds of the floor, u_{2} and v_{2} , are zero since it is assumed that the floor remains stationary during impact. The second time the ball hits the floor it has fallen from a height of 20 \mathrm{~m} . Its speed u_{1} is
u_{1}=\sqrt{2 g h}=\sqrt{2(9.8)(20)}=19.8 \mathrm{~m} / \mathrm{s}
By similar reasoning, the speed v_{1} of rebound may be found. The ball with speed v_{1} rises to a height of 14 \mathrm{~m} . Hence, its starting speed v_{1} is v_{1}=\sqrt{2(9.8)(14)}=16.6 \mathrm{~m} / \mathrm{s} . Now apply the following equation of impact, where downward speeds are chosen positive:
e=\frac{v_{2}-v_{1}}{u_{1}-u_{2}}=\frac{0-(-16.6)}{19.8-0}=\underline{0.84}
The value of e could be determined by considering the square roots of the successive heights to which the ball bounced. This is true since the value of e in this case is actually the ratio of the speeds which depend on the square roots of the heights e=\sqrt{14 / 20}=\underline{0.84} .