Q.
A particle is projected with a velocity u. So that its horizontal range and maximum height reached are equal. The maximum height reached is
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a
2u2/3g
b
4u2/5g
c
u2/g
d
8u2/17g
answer is D.
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Detailed Solution
R=Hmax2u2sinθcosθg=u2sin2θ2gtanθ=4sinθ=417Hmax=u2.16/172g=8u217g
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