A boy throws a ball into air at 45^ from the horizontal to land it on a roof of a building of height H . If the ball attains maximum height in 2 s and lands on the building in 3 s after launch, then value of H is _ _ _ _ m. ( g =10 ~m / s ² )

Step-by-step solution

For projectile motion at 45°, vertical component of initial velocity: v_ 0y = v₀ 45° = v₀ 2 Maximum height occurs at t = 2 s when v_y = 0 : 0 = v_ 0y - gt v_ 0y = 10 2 = 20 m/s Thus v₀ 2 = 20 , so v₀ = 20 2 m/s Height at landing time t = 3 s: H = v_ 0y t - 1 2 gt^2 = 20 3 - 1 2 10 9 = 60 - 45 = 15 m