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Two particles are projected from the same point with the same speed at different angles θ₁ and θ₂ to the horizontal. If their respective times of flights are T₁ and T₂ and horizontal ranges are same then
a) θ₁ + θ₂ = 90^o
b) T₁ = T₂ tan θ₁
c) T₁ = T₂ tan θ₂
d) T₁ sin θ₂ = T₂ sin θ₁

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a, b, d are correct

a, c, d are correct

b, c, d are correct

a, b, c are correct

answer is A.

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Detailed Solution

For complementary angles , ranges are same.

$∴ θ_{1} + θ_{2} = 90^{\circ}$

$T_{1} = \frac{2 usin θ_{1}}{g};$

$T_{2} = \frac{2 usin θ_{2}}{g} = \frac{2 usin (90^{o} - θ_{1})}{g} = \frac{2 ucos θ_{1}}{g}$

$\frac{T_{1}}{{sinθ}_{1}} = \frac{T_{2}}{{sinθ}_{2}} = \frac{T_{2}}{{cosθ}_{1}}$

$T_{1} = T_{2} \tan θ_{1}$

$T_{1} \sin θ_{2} = T_{2} \sin θ_{1}$

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