Irodov 1.319: The Optimal Hole Height – Amit & Deepali Sinha Foundation

Problem Irodov 1.319

A wide cylindrical vessel of height $H = 50$ cm, filled with water, rests on a table. Neglecting viscosity, a small hole is to be pierced in the vessel’s side wall. Find the height above the table at which the hole should be made so that the emerging water jet strikes the table at the greatest possible horizontal distance $l_{max}$ from the vessel, and find that maximum distance $l_{max}$ .

A side hole at height h drains the vessel; the jet falls a height h to the table and lands a distance l away. — A side hole at height $h$ drains the vessel; the jet falls a height $h$ to the table and lands a distance $l$ away.

Given

$H = 50$ cm

Find

Find

h

(the hole’s height above the table) that maximizes the range

l

, and find

l_{max}

Solution

Step 1Efflux speed from Torricelli’s law

The vessel is wide and open at the top, so the free surface falls negligibly slowly and the water above the hole is essentially static. With the hole a height $h$ above the table and the vessel filled to height $H$ , the depth of water above the hole is $H-h$ , so

v = \sqrt{2g(H-h)} \tag{1}

Step 2Time of fall and the range

The jet leaves horizontally and falls the same height $h$ down to the table, so the fall time is $t=\sqrt{2h/g}$ . Multiplying the exit speed by this fall time, with $v$ from (1), gives the horizontal range:

\begin{align*} l &= vt \\ &= \sqrt{2g(H-h)}\sqrt{\frac{2h}{g}} \\ &= 2\sqrt{h(H-h)}. \tag{2}\end{align*}

Step 3Maximize the range over the hole’s height

The range depends on $h$ only through the product $h(H-h)$ , so maximizing $l$ is the same as maximizing this product; setting its derivative with respect to $h$ to zero locates the maximum.

\frac{d}{dh}\bigl[h(H-h)\bigr] = \frac{d}{dh}\bigl[hH-h^2\bigr] = H-2h.

Setting this to zero and solving for $h$ ,

H - 2h = 0 \quad\Longrightarrow\quad h = \frac{H}{2}.

h = \frac{H}{2}

(3)

Step 4The maximum range

Substituting (3) into the range formula (2),

l_{max} = 2\sqrt{\frac{H}{2}\left(H-\frac{H}{2}\right)} = 2\sqrt{\frac{H^2}{4}} = 2\cdot\frac{H}{2}.

h = \dfrac{H}{2} = 25\ \text{cm}, \qquad l_{max} = H = 50\ \text{cm}

(4)

Check

At $h=0$ or $h=H$ the range is zero either way (no drop, or no head), so a maximum strictly between them is the only sensible outcome; the symmetric optimum $h=H/2$ and the clean result $l_{max}=H$ both confirm that.

Remark

Because $l$ depends on $h$ only through the symmetric product $h(H-h)$ , any two holes placed at complementary heights $h$ and $H-h$ land at the same distance: a low hole, with a strong squirt but only a short fall, and a high hole, with a weak squirt but a long fall, can strike the same spot on the table. The height $h=H/2$ is simply the one hole that is complementary to itself.