A boy throws a ball into air at $45^{\circ}$ from the horizontal to land it on a roof of a building

Question

A boy throws a ball into air at $45^{\circ}$ 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.
$$
\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\right)
$$

Professor PC Thomas Classes · Answer

The time to reach the maximum height is $t_{peak} = \frac{u_y}{g} = 2 	ext{ s}$, which implies the initial vertical velocity component is $u_y = 2 	imes 10 = 20 	ext{ m/s}$. The maximum height attained from the launch level is $H_{max} = \frac{u_y^2}{2g} = \frac{400}{20} = 20 	ext{ m}$. The ball lands on the roof at $t = 3 	ext{ s}$, meaning it has been falling from the peak for $\Delta t = 3 - 2 = 1 	ext{ s}$. The vertical distance fallen from the peak is $h_{fall} = \frac{1}{2}g(\Delta t)^2 = \frac{1}{2}(10)(1)^2 = 5 	ext{ m}$. The height of the building is $H = H_{max} - h_{fall} = 20 - 5 = 15 	ext{ m}$.