Two bodies are projected at angles θ and 90^o-θ to the horizontal with the same speed. The ratio of their times of flight is

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MARKS Content Team · Accepted Answer

Correct answer: (C) sin⁡θ:cos⁡θ. Time of flight, T=(2u sin⁡θ)/g For θ, T_1=(2u sin⁡θ)/g ......(i) For 90^o-θ, T_2=(2u sin⁡90^o-θ)/g=(2u cos⁡θ)/g .....(ii) Dividing Equation. (i) by Equation. (ii), we get ∴ T_1/T_2=(2u sin⁡θ)/g×g/(2u cos⁡θ)=(sin⁡θ)/(cos⁡θ)

Two bodies are projected at angles $θ$ and $(90^{o} - θ)$ to the horizontal with the same speed. The ratio of their times of flight is

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More Motion In Two Dimensions Questions

Two bodies are projected at angles θ and 90o-θ to the horizontal with the same speed. The ratio of their times of flight is

Solution

More Motion In Two Dimensions Questions

Two bodies are projected at angles $θ$ and $(90^{o} - θ)$ to the horizontal with the same speed. The ratio of their times of flight is