8) a rubber ball is dropped from a height of 5 m on a plane where the acceleration dure to gravity is not
known on bouncing in rises to a height of 1.8 m. The ball loses its velocity on bouncing by a factor of
a. 16/25
c. 315
d. 9/25
Answers
Answered by
1
Explanation:
A ball dropped from a height h on reaching the surface velocity is given by
V1= √2gh1
Let V2 be the velocity with which the ball bounces. It will attain a height h2 given by
V2= √2gh2
V2/V1 = √h2/h1
Given
h1 = 5 m h2 = 1.8 m
V2/V1 = √1.8/5 = √0.36 = 0.6
1 - V2/V1 = V1 - V2 / V1 = 1 - 0.6/1
ANS) 0.4 = 2/5
Similar questions