Two projectiles are fired from the ground with the same initial speed, and they cover the same horizontal range. The ratio of their minimum speeds during flight is 1 : 3 . If the sum of their times of flight is 8 s, the maximum height reached by the projectile that has the longer time of flight is (Take g = 10 m/s ^2 )

Step-by-step solution

For two projectiles to have the same horizontal range with the same initial speed u , their angles of projection must be and 90^ - . The minimum speed of a projectile occurs at the highest point of its trajectory and is equal to the horizontal component of its velocity, v_ min = u . The ratio of their minimum speeds is given as 1 : 3 . Assuming is the steeper angle (so is smaller): u u (90^ - ) = = = 1 3 This implies = 3 . The times of flight are t₁ = 2u g and t₂ = 2u g . The ratio of their times of flight is: t₁ t