Question 14: Projectiles from a Window
Solution
Let the height of the window be $h$. Pebbles are thrown with speed $u$ in all directions. We need to ensure that regardless of the throw direction, the angle of impact $\phi$ with the ground is always greater than or equal to a critical angle $\theta$.
Analysis:
The angle of impact $\phi$ is given by $\tan \phi = \frac{|v_y|}{v_x}$.
To satisfy $\phi \ge \theta$, we must have $\tan \phi \ge \tan \theta$.
This implies we need to find the minimum possible impact angle. The impact angle is minimized when the horizontal velocity component $v_x$ is maximized (since $v_y$ is determined by energy conservation and $v_x$, maximizing $v_x$ minimizes the ratio $|v_y|/v_x$).
The maximum horizontal velocity at impact is equal to the maximum horizontal projection velocity, which occurs when the pebble is thrown horizontally.
Case: Horizontal Projection ($u_x = u, u_y = 0$)
At impact:
$$ v_x = u $$
$$ v_y = \sqrt{2gh} $$
The angle of impact $\phi_{\min}$ in this limiting case is:
$$ \tan \phi_{\min} = \frac{v_y}{v_x} = \frac{\sqrt{2gh}}{u} $$
Condition:
The problem states the pebbles hit at an angle $\theta$ or greater. Thus, the minimum angle must be $\theta$.
$$ \tan \theta = \frac{\sqrt{2gh}}{u} $$
Squaring both sides:
$$ \tan^2 \theta = \frac{2gh}{u^2} $$
Solving for $h$:
$$ h = \frac{u^2 \tan^2 \theta}{2g} $$
Answer: $$ h = \frac{u^2 \tan^2 \theta}{2g} $$