A particle is projected with a velocity , so that its range on a horizontal plane is twice the greatest height attained. If g is acceleration due to gravity, then its range is
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a
4u25g
b
4g5v2
c
4u35g2
d
4u5g2
answer is A.
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Detailed Solution
H=u2sin2θ2g and R=u2sin2θg Since,R=2H, so u2sin2θg=2u2sin2θ2g or 2sinθcosθ=sin2θ or tanθ=2 ∴R=u22gsinθcosθ=2u2g2515=4u25g.