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Question

A particle having mass m is projected with a velocity v0 from a point P on a horizontal ground making an angle θ with horizontal. Find out the torque about the point of projection acting on the particle when it is at its maximum height?


A

mv20 sin 2θ4

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B

mv20 sin 2θ2

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C

mv20 sin2θ2

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D

2mv20 sin 2θ2

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Solution

The correct option is B

mv20 sin 2θ2


τ = F r sin θ

As r sin θ is nothing but the horizontal component of r which is half of the Range, R.
τ=Fr sinθ=mgR2=v20sin2θ2gmg
τ=mv20sin2θ2


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