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Question

A body of mass m is projected from ground with speed u at an angle θ with horizontal. The power delivered by gravity when it is at half of maximum height from ground is

A
mgucosθ2
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B
mgusinθ2
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C
mgusinθ2
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D
Both (b) and (c)
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Solution

The correct option is D Both (b) and (c)

Hmax=u2sin2θ2g
Applying 3rd eqn of motion at half of maximum height :
v2y=(usinθ)22g(Hmax2)
v2y=(usinθ)22g u2sin2θ4g
vy=±usinθ2
for the two points of half the maximum height

At point A, P=F.V
=(mg)(usinθ2)cosπ=mgusinθ2
{angle between vectors is 180}

At point B,
P=F.V {angle between vectors here is 0}
P=+mgusinθ2

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