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A particle is projected with a velocity v, so that its horizontal range twice the greatest height attained. The horizontal range is
detailed solution
Correct option is A
Given, a particle having horizontal range is twice the greatest height attained by it, i.e. R = 2Hv2sin2θg=2v2sin2θ2g ⇒2sinθcosθ=sin2θ⇒tanθ=2⇒ sinθ=25⇒cosθ=15∴ R=v2sin2θg=v2×2sinθcosθg⇒ R=v2×2g×25×15 i.e. R=4v25gTalk to our academic expert!
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