a body falling from a height of 10m rebounds from a hard floor.It loses 20% of its energy in the impact.What could be the height to which it would rise after the impact
Asked by Ashwnikumar | 5th Dec, 2015, 06:35: PM
Expert Answer:
Given that the body is dropped from a height, h1 = 10 m
Potential energy at h1 = mgh1
Given g =10 m/s2 , h1 = 10 m
Therefore the potential energy at h1 = m×10×10
= 100m J
On striking the ground level, the ball looses 20% of its initial energy:
i.e. (20/100)× 100m J
=20m J
The energy left on striking the ground = 100 m J - 20 m J
= 80 m J
So the final energy of the ball , mgh2 = 80m J
i.e h2 = (80 /10)
= 8 m
Hence the body will bounce back to a height of 8 m after the impact.
Given that the body is dropped from a height, h1 = 10 m
Potential energy at h1 = mgh1
Given g =10 m/s2 , h1 = 10 m
Therefore the potential energy at h1 = m×10×10
= 100m J
On striking the ground level, the ball looses 20% of its initial energy:
i.e. (20/100)× 100m J
=20m J
The energy left on striking the ground = 100 m J - 20 m J
= 80 m J
So the final energy of the ball , mgh2 = 80m J
i.e h2 = (80 /10)
= 8 m
Hence the body will bounce back to a height of 8 m after the impact.
Answered by Yashvanti Jain | 6th Dec, 2015, 09:58: AM
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