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

Question

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 (A) sin θ: 1 (B) cos θ: 1 (C) sin θ :cos ⁡ θ  (D) cos θ :sin ⁡ θ

Tardigrade Admin · Accepted Answer

Time of flight, T=(2 u sin θ /g) For θ , T1=(2 u sin θ /g) ......(i) For (90o - θ ), T2=(2 u sin ((90)o - θ )/g)=(2 u cos â¡ θ /g) .....(ii) Dividing Equation. (i) by Equation. (ii), we get ∴ (T1/T2)=(2 u sin θ /g)× (g/2 u cos â¡ θ )=(sin â¡ θ /cos â¡ θ )

Q. 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

$s in θ : 1$

$cos θ : 1$

$s in θ : cos θ$

$cos θ : s in θ$

Q. 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

sinθ:1

cosθ:1

sinθ:cos⁡θ

cosθ:sin⁡θ

Time of flight, T=g2usinθ​ For θ,T1​=g2usinθ​ ......(i) For (90o−θ),T2​=g2usin((90)o−θ)​=g2ucos⁡θ​ .....(ii) Dividing Equation. (i) by Equation. (ii), we get ∴T2​T1​​=g2usinθ​×2ucos⁡θg​=cos⁡θsin⁡θ​

Q. 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

$s in θ : 1$

$cos θ : 1$

$s in θ : cos θ$

$cos θ : s in θ$