question_answer

A ball is dropped from a height of 5 m onto a sandy floor and penetrates the sand up to 10cm before coming to rest. What is the retardation of the ball in sand assuming it to be uniform?

Answer:

Velocity on reaching the top of the sandy floor, V \[=\sqrt{2gh}=\sqrt{2\times 9.8\times 5}\] \[=\sqrt{98}\,m/s\]                         ??????.(1) at the top of the sandy floor, the final velocity becomes initial velocity and final velocity of the ball is zero. from (1), using \[{{v}^{2}}-{{u}^{2}}=2as,\] we get, \[0-98=2\times a\times 10\times {{10}^{-2}}\] \[a=-490\,\,m/{{s}^{2}}\]