a rubber ball is dropped from a height of 5 m on a plane where the acceleration due to gravity is not known. on bouncing it rises to a heifht of 1.8 m. the ball loses its velocity on bouncing by a factor of?
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Answered by
302
HEY!!
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
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
Answered by
14
Answer:
The ball loses its velocity on bouncing by a factor of 2/5
Explanation:
The following image explains it,
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