An airplane and a ship move in the same vertical plane along…

An airplane and a ship move in the same vertical plane along parallel straight lines: the ship with speed $v_1$, the airplane with speed $v_2$, at a height $h$ above the ship. At what horizontal distance from the ship must the bomb be released to hit it, if (a) they move in the same direction; (b) they move in opposite directions?

Explanation of Method

Bombing a moving ship.

Step 1: Fall time

\[ h = \tfrac12 g t^2 \;\Longrightarrow\; t = \sqrt{\frac{2h}{g}}. \]

Step 2: Look from the ship

The bomb keeps the plane's horizontal speed $v_2$. Relative to the ship it moves horizontally with the difference or the sum of the speeds, and it must cover the release distance $D$ in the fall time:

\[ D = v_{\text{rel}}\,t. \]

$\text{(a) same direction}$ $v_2$ $v_1$ $D = (v_2 - v_1)\sqrt{2h/g}$ $h$ $\text{(b) opposite directions}$ $v_2$ $v_1$ $D = (v_2 + v_1)\sqrt{2h/g}$ $h$

At release (solid ship) and at impact (dashed ship). While the bomb falls for $t = \sqrt{2h/g}$, it moves forward like the plane and the ship moves too; the release distance $D$ is what the bomb gains on the ship in that time.

Step 3: (a) Same direction

\[D_a = (v_2 - v_1)\sqrt{\frac{2h}{g}}\]

Step 4: (b) Opposite directions

\[D_b = (v_2 + v_1)\sqrt{\frac{2h}{g}}\]

Result

\[\boxed{\,D_a = (v_2 - v_1)\sqrt{\frac{2h}{g}}\,}\]

\[\boxed{\,D_b = (v_2 + v_1)\sqrt{\frac{2h}{g}}\,}\]

Remark

Note. The distance is measured horizontally between the plane and the ship at the moment of release, with the ship ahead of the plane. For a numerical example, $h = 1000\ \mathrm{m}$, $v_2 = 150\ \mathrm{m/s}$, $v_1 = 10\ \mathrm{m/s}$: $t \approx 14.3\ \mathrm{s}$, $D_a \approx 2.0\ \mathrm{km}$, $D_b \approx 2.3\ \mathrm{km}$.