Irodov 1.135: Optimal Shelf Height Maximizes Landing Range

Problem Irodov 1.135

A small disc $A$ starts from rest at the top of a smooth hill of total height $H$ . The hill’s profile descends smoothly and then levels out into a horizontal shelf at some height $h$ above the ground, $0<h<H$ , before the shelf ends; the disc leaves the shelf’s edge and falls freely as a projectile, landing at a horizontal distance $s$ measured from the point on the ground directly below that edge. Find the height $h$ of the horizontal portion that makes the landing distance $s$ as large as possible, and find that maximum distance.

The hill profile: the disc starts from rest at $A$ , height $H$ ; the smooth track flattens into a shelf at height $h$ , and from its edge the disc falls as a projectile (dashed) through the horizontal distance $s$ .

Given

$H$
smooth hill, disc released from rest at the top

Find

h

that maximizes

s

, and the resulting

s_{\max}

Solution

Step 1Speed on the shelf from energy conservation

The track is smooth, so mechanical energy is conserved from the release point down to the shelf, a drop in height of $H-h$ .

\begin{align*}\frac{1}{2}mv^2 &= mg(H-h) \\ v^2 &= 2g(H-h)\end{align*}

v = \sqrt{2g(H-h)}

(1)

Step 2Time of fall from the shelf’s edge

Leaving the edge with a purely horizontal velocity, the disc falls the remaining height $h$ under gravity alone.

\begin{align*}h &= \frac{1}{2}gt^2 \\ t^2 &= \frac{2h}{g}\end{align*}

t = \sqrt{\dfrac{2h}{g}}

(2)

Step 3Range as a function of the shelf height

The horizontal range is the shelf speed (1) times the fall time (2).

\begin{align*} s(h) &= v\,t \\ &= \sqrt{2g(H-h)}\,\sqrt{\dfrac{2h}{g}} \\ &= \sqrt{4h(H-h)} \\ &= 2\sqrt{h(H-h)} \tag{3}\end{align*}

Step 4Maximizing the range over

h

Since $s(h)$ from (3) is largest exactly where the product $h(H-h)$ is largest, differentiate that product with respect to $h$ and set the result to zero.

\begin{align*}\frac{d}{dh}\bigl[h(H-h)\bigr] &= H-2h \\ 0 &= H-2h\end{align*}

h = \dfrac{H}{2}

(4)

This shelf sits exactly halfway up the hill.

Step 5The maximum range

Substituting $h=H/2$ into the range formula (3) gives the maximum distance.

\begin{align*}s_{\max} &= 2\sqrt{\dfrac{H}{2}\left(H-\dfrac{H}{2}\right)} \\ &= 2\sqrt{\dfrac{H}{2}\cdot\dfrac{H}{2}}\end{align*}

s_{\max} = H

(5)

Check

At either extreme the range vanishes, as it must: $h\to0$ gives full speed but zero fall time, and $h\to H$ gives zero speed but full fall time, and both make $s\to0$ in (3); the maximum found at $h=H/2$ , sitting midway between these two zeros, is therefore the single interior peak, not an endpoint artifact.

Remark

Step 4 is really an arithmetic-mean/geometric-mean statement in disguise: for two nonnegative numbers with a fixed sum $H$ , their product $h(H-h)$ is largest exactly when the two are equal, $h=H-h$ . The same trick maximizes a rectangle’s area for a fixed perimeter, or a projectile’s range for a fixed launch speed at 45 degrees, whenever the quantity to optimize factors into two terms that must add to a constant.