A projectile fired at 30 ° to the ground is observed to be at same height at time 3   s and 5   s after projection, during its flight. The speed of projection of the projectile is m   s - 1 . (Given g = 10   m   s - 2 )

Step-by-step solution

The time taken to reach maximum height is T ' = u sin θ g . The total time of flight is T = ( 3 + 5 )   s = 8   s . Hence, T ' = T 2 = 3 + 5 2 = u sin θ g ⇒ 4 = ( u ) × 1 2 10 ⇒ u = 80   m   s - 1