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Question

A ball is dropped from a height of 10 m. If the energy of the ball reduces by 40% after striking the ground, how high will the ball bounce back after striking the ground for the first time (in meters)?
Take g=10 m/s2

A
6
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B
4
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C
3
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D
9
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Solution

The correct option is A 6
Given:
Initial height of the ball, h=10 m
Acceleration due to gravity, g=10 m/s2

Let the mass of the object be m.

Initially, the ball has only potential energy stored in it. So, total energy of the ball will be:
E=U
Using formula of potential energy:
U=mgh

E=mgh

After hitting to the ground, 40% energy of the ball reduces. In other words, 60% of energy is retained. Let the energy left be Ef.
EfE=60100
Ef=0.6E

After bouncing, again at the highest point, the entire energy of the ball is in the form of potential energy. Let the final height of the ball be hf. Using this and conservation of energy,
Ef=mghf
0.6E=mghf
0.6mgh=mghf
hf=0.6h
hf=0.6×10=6 m

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