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Question

A particle is projected at an angle θ from the horizontal with kinetic energy K. What is the kinetic energy of the particle at the highest point?

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Solution

Given, Kinetic energy K, Let velocity of particle is v.
12mv2=K v=2Km
Velocity in horizontal direction=vcos(θ)=2Kmcos(θ)
Velocity in vertical direction=vsin(θ)=2Kmsin(θ)
At highest point, Vertical velocity is zero only constant horizontal velocity is present.
Kinetic energy at highest point =12m(vcos(θ))2=12×m×2Kmcos2(θ)=Kcos2(θ)

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