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Question

A body of 1 kg is thrown vertically upwards with an initial velocity (u)=2 m/s, the height at which kinetic energy of the body is (14)th of its original value is (take g=10 m/s2)

A
0.15 m
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B
0.25 m
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C
0.3 m
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D
0.35 m
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Solution

The correct option is A 0.15 m
Given, initial velocity (u)=2 m/s
Mass of body (m)=1 kg
Initial kinetic energy (Ki)=12mu2=12×1×(2)2=2 J
Final kinetic energy (Kf)=14 Initial kinetic energy (Ki)
From the law of conservation of energy,
Ui+Ki=Uf+Kf
Ui=0 J;Uf=mgh J
Ki=Uf+Kf
Uf=KiKf
=KiKi4=3Ki4=34×2=32 J
mgh=32 J
h=32×mg=32×1×10=320=0.15 m

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