Two particles are projected from the same point with the same speed at different angles θ1 and θ2 to the horizontal. They have the same range. Their times of flight are t1 and t2 respectively.
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a
t1t2=tan2θ1
b
t1sinθ1=t2cosθ2
c
t1t2=tanθ1
d
t1t2=tan2θ2
answer is C.
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Detailed Solution
R=u2sin2θg is same for angles θ&90∘−θt1=2usinθ1g=2usinθ1gt2=2usinθ2g=2usin90∘−θ1gt1t2=sinθ1sin90∘−θ1=sinθ1cosθ1=tanθ1