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Question

The velocity at the maximum height of a projectile is half its initial velocity of projection. Find its range on the horizontal plane. If u=20 ms1. (Take g=10 m/s2)

A
203 m/s
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B
103 m/s
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C
15 m/s
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D
10 m/s
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Solution

The correct option is A 203 m/s
We know that velocity at the maximum height =ucosθ
According to question
ucosθ=u2
θ=60
Range of a projectile =R=u2sin2θg
R=u2sin(120)g=400×3210=203 m

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