What is the smallest speed with which a stone must be…

What is the smallest speed with which a stone must be thrown, from a height $h$, to get over a rectangular obstacle of width $l$ and height $H$ ($H \gt h$)? The thrower may choose both the angle and the distance from the obstacle.

Explanation of Method

Smallest speed to clear a rectangular obstacle.

Step 1: Speed depends only on height

For any projectile, $v^2 = v_0^2 - 2g\,(y - y_0)$: the speed at a given height does not depend on the direction of motion. At the roof height $H$, a stone thrown from $h$ has

\[ u^2 = v_0^2 - 2g\,(H - h). \]

Step 2: Crossing the roof

Above the roof the stone must get from one top corner to the other, a horizontal distance $l$ at the same height. That is a throw with range $l$ starting with speed $u$, and the smallest speed for range $l$ is at 45° (Gravitational Kinematics 045):

\[ \frac{u^2}{g} \ge l \;\Longrightarrow\; u_{\min}^2 = g\,l. \]

Step 3: Minimum launch speed

\[ v_0^2 = u_{\min}^2 + 2g\,(H - h) = g\,(l + 2H - 2h). \]

\[v_{0,\min} = \sqrt{g\,(l + 2H - 2h)}\]

$l$ $H$ $h$ $\text{grazes both corners}$ $45^\circ$ $v_{\min} = \sqrt{g\,(l + 2H - 2h)}$

The cheapest throw just grazes both top corners and crosses the roof like a 45° throw of range $l$ (drawn for $l = 4$, $H = 3$, $h = 1$ m). The speed needed at the corners is $\sqrt{gl}$; rising the extra $H - h$ costs the rest.

Result

\[\boxed{\,v_{0,\min} = \sqrt{g\,(l + 2H - 2h)}\,}\]

Remark

Note. The best throw grazes both top corners, passing them at 45°. The original answer is correct, but it was stated without a derivation ("turns out to be"); the two steps above show where it comes from. Example: $l = 4\ \mathrm{m}$, $H = 3\ \mathrm{m}$, $h = 1\ \mathrm{m}$ gives $v_{0,\min} = \sqrt{9.81 \cdot 8} \approx 8.9\ \mathrm{m/s}$.