The initial speed of a projectile fired from ground is u . At the highest point during its motion, the speed of projectile is 3 2 u . The time of flight of the projectile is:

Step-by-step solution

At maximum height, only horizontal velocity of the particle remains. Let the angle of project be θ . The horizontal velocity of the particle will remain same, therefore u cos θ = 3 u 2 ⇒ θ = 30 ° Now, time of flight of projectile is T = 2 u sin θ g = u g