At what height does a shell hit a vertical wall x = 4 km…

At what height does a shell hit a vertical wall $x = 4\ \mathrm{km}$ from the cannon? The shell is fired at $v_0 = 400\ \mathrm{m/s}$ and $\alpha = 14^\circ$ to the horizontal (no air resistance).

Explanation of Method

Where a shell hits a wall.

Step 1: Time to reach the wall

\[ t = \frac{x}{v_0\cos\alpha} = \frac{4000}{400 \cdot 0.9703} \approx 10.31\ \mathrm{s}. \]

Step 2: Height at that time

\[ y = x\tan\alpha - \frac{g x^2}{2v_0^2\cos^2\alpha} = 4000 \cdot 0.24933 - \frac{9.81 \cdot 1.6 \cdot 10^7}{2 \cdot 160\,000 \cdot 0.94147}. \]

\[ y = 997.3 - \frac{1.5696 \cdot 10^8}{301\,270} = 997.3 - 521.0. \]

\[y \approx 476\ \mathrm{m}\]

Check with time

$ y = v_0\sin\alpha\,t - \tfrac12 g t^2 = 96.77 \cdot 10.306 - 4.905 \cdot 10.306^2 $

$ y = 997.3 - 521.0 \approx 476.3\ \mathrm{m}\ \checkmark $

$476\ \mathrm{m}$ $4\ \mathrm{km}$ $\text{wall}$ $v_0 = 400\ \mathrm{m/s},\ 14^\circ$ $\text{top: } 477\ \mathrm{m}$

Vertical scale stretched about 6×. The top of the path (477 m) is at 3.83 km, just before the wall, so the shell hits the wall only 1 m below its highest point. The dashed part shows where it would have gone without the wall.

Result

\[\boxed{\,y \approx 476\ \mathrm{m}\,}\]

Remark

About the original solution. The method was right, but $2 \cdot 160\,000 \cdot 0.9415$ is 301 280, not 300 480, which gave 474.6 m instead of about 476 m. Interestingly, the highest point of the path (477 m) is at 3.83 km, so the shell hits the wall just past its top.