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Question

A projectile is thrown at an angle of 60 with horizontal. If E is the kinetic energy with which the projectile is thrown, then find the kinetic energy at the top of trajectory.

A
3E4
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B
E4
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C
E2
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D
3E2
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Solution

The correct option is B E4
Let u be the velocity with which the projectile is thrown.
We know that kinetic energy is given as
E=12mu2
When the projectile is at the top of its trajectory then its velocity is minimum. Only horizontal component of velocity exists at the top of its trajectory .
vtop=ucosθ (where θ is the angle of projection with the horizontal)
Kinetic energy at the top of trajectory Et=12m(ucosθ)2
Et=12m(ucos60)2
Et=E4

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