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Question

A particle is projected with a velocity v so that its horizontal range twice the greatest height attained. The horizontal range is:



A

4v25g

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B

v2g

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C

v22g

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D

2v23g

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Solution

The correct option is A

4v25g


R=V2 sin 2θg=2×V2 sinθg

H=(V2sin2θ2g)    

Given, R=2H

2sinθ cosθ=sin2θ
tanθ=2
sinθ=25     cosθ=15
R=V2 sin2θg=V2×2 sinθ cosθg=V2×2g×25×15
R=4V25g


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