Two projectiles are thrown with the same initial speed at two different angles of projection such that they cover the same horizontal range of 120 m. If the sum of the squares of their times of flight is 52 s ^2 , then the absolute difference between their times of flight is (Take g = 10 m/s ^2 )

Step-by-step solution

For two projectiles to cover the same horizontal range R with the same initial speed u , their angles of projection must be complementary, i.e., and 90^ - . The times of flight for these two angles are: t₁ = 2u g t₂ = 2u g The product of their times of flight is: t₁ t₂ = ( 2u g ) ( 2u g ) = 2 g ( u^2 (2 ) g ) = 2R g Given R = 120 m and g = 10 m/s ^2 : t₁ t₂ = 2 120 10 = 24 s ^2 We are given the sum of the squares of the times of flight: t₁^2 + t₂^2 = 52 s ^2 Using the algebraic identity for the square of a differen