A tank drives straight toward a cannon at a constant v = 3.6…

A tank drives straight toward a cannon at a constant $v = 3.6\ \mathrm{km/h} = 1\ \mathrm{m/s}$. With what initial speed must a shell be fired at $\alpha = 30^\circ$ to hit the tank, if the tank was $d = 8\ \mathrm{km}$ away at the moment of firing?

Explanation of Method

Hitting an approaching tank.

Step 1: Where the shell and the tank meet

The shell lands after $t = \dfrac{2v_0\sin\alpha}{g} = \dfrac{v_0}{g}$ (since $\sin 30^\circ = \tfrac12$). By then the tank has come $v t$ closer: $v_0\cos\alpha \cdot t = d - v t$.

Step 2: Quadratic for $v_0$

\[ \frac{v_0^2\sin 60^\circ}{g} = d - \frac{v\,v_0}{g}, \]

\[ 0.8660\,v_0^2 + 1 \cdot v_0 - 78\,480 = 0. \]

\[ v_0 = \frac{-1 + \sqrt{1 + 4 \cdot 0.8660 \cdot 78\,480}}{2 \cdot 0.8660} = \frac{-1 + 521.4}{1.732}. \]

\[v_0 \approx 300.5\ \mathrm{m/s}\]

Check

Flight time $ t = 300.46/9.81 \approx 30.63\ \mathrm{s} $; shell range $ 300.46 \cdot 0.866 \cdot 30.63 \approx 7969\ \mathrm{m} $

Tank position $ 8000 - 1 \cdot 30.63 \approx 7969\ \mathrm{m}\ \checkmark $

$d = 8\ \mathrm{km} \text{ at firing}$ $\text{tank, } 1\ \mathrm{m/s}$ $v_0 \approx 300.5\ \mathrm{m/s},\ 30^\circ,\ t \approx 30.6\ \mathrm{s}$

Drawn to scale. During the 30.6 s flight the tank drives only about 31 m closer (dashed), so the required speed is just slightly less than for a fixed target at 8 km.

Result

\[\boxed{\,v_0 \approx 300.5\ \mathrm{m/s}\,}\]

Remark

About the original solution. Its equation is correct, but the stated solution 277.55 m/s does not satisfy it: with 277.55 m/s the shell would land only about 6.8 km away. The correct value is about 300.5 m/s. Because the tank is slow, this is barely less than the 301.0 m/s needed for a fixed target at 8 km.