Question

a projectile can have the same range for two diff angles of projection if t1 and t2 are time of flight in two cases then product of t1 and t2 is proportional to

Answer 1

The two angles having the same range will be complementary which is Q and 90-Q

T1 = 2u sinQ/g
T2= 2u cosQ/g

T1T2= 4u square sinQcosQ/g square
T1T2= 2R/g
TiT2 is proportional to R.

Answer 2

Assuming the initial speed u is same.
Range R = u² Sin 2Ф₁ / g  =  u² Sin 2Ф₂ / g 
      So Sin 2Ф₁  = Sin 2Ф₂    =>   2Ф₂ = 180° - 2Ф₁
           Ф₂ = 90° - Ф₁

Times of flight are t₁ and t₂.
      So R = u Cos Ф₁ * t₁ = u Cos Ф₂ * t₂
                                        = u Sin Ф₁ * t₂

 t₁ * t₂ =  R² / [u² Cos Ф₁ Sin Ф₁ ]
          = 2 R² / [u² Sin 2Ф₁ ]
          = 2 R² / [ R * g ]
          = 2 R / g

So answer is the option (4)  R...