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Question

The speed of a projectile when it is at its greatest height is 2/5 times its speed at half the maximum height. What is its angle of projection?

A
30o
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B
60o
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C
45o
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D
0o
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Solution

The correct option is B 60o
Time taken to reach max height =usinθg Max height =u2sin2θ2g
Half the max height =u2sin2θ4g Horizontal 'v' here ucosθ
Vertical velocity at half max height

=u2sin2θ2g(usinθ2g)=usinθ2

Velocity at half the max-height

=u2cos2θ+u2sin2θ2ucosθ=25u2cos2θ+u2sin2θ2u2cos2θ=25u2(1sin2θ+sin2θ2)5(1sin2θ)=2sin2θ3=4sin2θsinθ=32=θ=π3(60)
Option B

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