Question

A projectile can have the same range R for two angles of projection. If t₁ & t₂ be the times of flight in the two cases then what is the product of two times of flight ?

(a) t₁t₂ ∝ R².

(b) t₁t₂ ∝ R. 

(c) t₁t₂ ∝ 1/R. 

(d) t₁t₂ ∝ 1/R².

Answer 1

Answer is (b) t₁t₂ ∝ R.

Range is same for angles of projection θ and (90 - θ)

t₁ = 2u sinθ/g

t₂ = [2u sin(90 - θ)]/ g = 2u cosθ/g. 

hence t₁t₂ = 4(u² sinθ cosθ/g²) = 2/g[u² sinθ/g]

= (2/g)R., Where R is range.

hence t₁t₂ ∝ R because R is constant.