A rubber ball is dropped from a height of $20 \mathrm{~m}$ onto a hard plane. Upon bouncing, it rise

Question

A rubber ball is dropped from a height of $20 \mathrm{~m}$ onto a hard plane. Upon bouncing, it rises to a maximum height of $5 \mathrm{~m}$. The fractional loss in its velocity due to the bounce is (Take $g = 10 \mathrm{~m/s}^2$)

Professor PC Thomas Classes · Answer

(b) : The velocity of the ball just before striking the ground is $v_i = \sqrt{2gh_1} = \sqrt{2 	imes 10 	imes 20} = \sqrt{400} = 20 \mathrm{~m/s}$. The velocity of the ball just after rebounding is $v_f = \sqrt{2gh_2} = \sqrt{2 	imes 10 	imes 5} = \sqrt{100} = 10 \mathrm{~m/s}$. The magnitude of velocity lost is $20 - 10 = 10 \mathrm{~m/s}$. The fractional loss in velocity is $\frac{	ext{Loss}}{	ext{Initial}} = \frac{10}{20} = \frac{1}{2}$.