Two airplanes fly one behind the other along the same line…

Two airplanes fly one behind the other along the same line, $d = 1080\ \mathrm{m}$ apart, both with speed $v_1 = 1200\ \mathrm{km/h}$ relative to the ground. A shell is fired from the rear airplane toward the front one with speed $v_2 = 300\ \mathrm{m/s}$ relative to the airplane. After what time does the shell hit the front airplane? How far does the airplane travel in this time?

Explanation of Method

A shell fired between two airplanes.

Step 1: Convert the airplane speed

\[ v_1 = \frac{1200 \cdot 1000\ \mathrm{m}}{3600\ \mathrm{s}} = \frac{1000}{3}\ \mathrm{m/s} \approx 333.3\ \mathrm{m/s}. \]

Step 2: Look from the airplanes

Both airplanes have the same velocity, so relative to each other they are at rest and the gap stays $d = 1080\ \mathrm{m}$. The shell's speed is given relative to the airplane, so in the airplanes' frame it simply crosses the gap at $v_2$:

\[ t = \frac{d}{v_2} = \frac{1080\ \mathrm{m}}{300\ \mathrm{m/s}}. \]

\[t = 3.6\ \mathrm{s}\]

Step 3: Distance flown by the airplane

\[ s = v_1\,t = \frac{1000}{3}\ \mathrm{m/s} \cdot 3.6\ \mathrm{s}. \]

\[s = 1200\ \mathrm{m}\]

Check from the ground

shell speed relative to the ground: $ v_1 + v_2 = 333.3 + 300 = 633.3\ \mathrm{m/s} $ shell path: $ 633.3 \cdot 3.6 = 2280\ \mathrm{m} = 1200\ \mathrm{m} + 1080\ \mathrm{m}\ \checkmark $

$t = 0$ $t = 3.6\ \mathrm{s}$ $\text{shell}$ $d = 1080\ \mathrm{m}$ $\text{airplane: } 1200\ \mathrm{m}$ $\text{shell: } 2280\ \mathrm{m}$ $\text{front}$ $\text{rear}$

Seen from the ground: while the shell flies, both airplanes move 1200 m forward. The shell covers the airplane's 1200 m plus the 1080 m gap, i.e. 2280 m.

Result

\[\boxed{\,t = 3.6\ \mathrm{s}\,}\]

\[\boxed{\,s = 1200\ \mathrm{m}\,}\]

Remark

Note. The speed of the airplanes does not change the time at all (air resistance is ignored). It is like throwing a ball to a friend inside a moving train: only the speed relative to the train matters.