Solution 17
1. Coordinates of the Bunker
The bunker is located at distance $d$ and angle $\phi$. $$ x_B = d \cos\phi $$ $$ y_B = d \sin\phi $$
The bunker is located at distance $d$ and angle $\phi$. $$ x_B = d \cos\phi $$ $$ y_B = d \sin\phi $$
2. Coordinates of the Shell
Let the firing angle be $\alpha$. At time $T$, the shell coordinates are: $$ x_S = u T \cos\alpha $$ $$ y_S = u T \sin\alpha – \frac{1}{2}gT^2 $$
Let the firing angle be $\alpha$. At time $T$, the shell coordinates are: $$ x_S = u T \cos\alpha $$ $$ y_S = u T \sin\alpha – \frac{1}{2}gT^2 $$
3. Minimizing the Separation
To explode closest to the bunker, we must minimize the square of the distance $D^2$ between the shell and the bunker: $$ D^2 = (x_S – x_B)^2 + (y_S – y_B)^2 $$ $$ D^2 = (u T \cos\alpha – x_B)^2 + (u T \sin\alpha – (y_B + \frac{1}{2}gT^2))^2 $$ Let $K = y_B + \frac{1}{2}gT^2 = d\sin\phi + \frac{1}{2}gT^2$. We differentiate $D^2$ with respect to $\alpha$ and set to zero: $$ \frac{d(D^2)}{d\alpha} = 2(u T \cos\alpha – x_B)(-u T \sin\alpha) + 2(u T \sin\alpha – K)(u T \cos\alpha) = 0 $$ Dividing by $2 u T$: $$ -\sin\alpha (u T \cos\alpha – x_B) + \cos\alpha (u T \sin\alpha – K) = 0 $$ $$ -u T \sin\alpha \cos\alpha + x_B \sin\alpha + u T \sin\alpha \cos\alpha – K \cos\alpha = 0 $$ $$ x_B \sin\alpha – K \cos\alpha = 0 $$ $$ \tan\alpha = \frac{K}{x_B} $$
To explode closest to the bunker, we must minimize the square of the distance $D^2$ between the shell and the bunker: $$ D^2 = (x_S – x_B)^2 + (y_S – y_B)^2 $$ $$ D^2 = (u T \cos\alpha – x_B)^2 + (u T \sin\alpha – (y_B + \frac{1}{2}gT^2))^2 $$ Let $K = y_B + \frac{1}{2}gT^2 = d\sin\phi + \frac{1}{2}gT^2$. We differentiate $D^2$ with respect to $\alpha$ and set to zero: $$ \frac{d(D^2)}{d\alpha} = 2(u T \cos\alpha – x_B)(-u T \sin\alpha) + 2(u T \sin\alpha – K)(u T \cos\alpha) = 0 $$ Dividing by $2 u T$: $$ -\sin\alpha (u T \cos\alpha – x_B) + \cos\alpha (u T \sin\alpha – K) = 0 $$ $$ -u T \sin\alpha \cos\alpha + x_B \sin\alpha + u T \sin\alpha \cos\alpha – K \cos\alpha = 0 $$ $$ x_B \sin\alpha – K \cos\alpha = 0 $$ $$ \tan\alpha = \frac{K}{x_B} $$
4. Final Calculation
Substitute back $x_B$ and $K$: $$ \tan\alpha = \frac{d\sin\phi + \frac{1}{2}gT^2}{d\cos\phi} $$ $$ \tan\alpha = \frac{d\sin\phi}{d\cos\phi} + \frac{gT^2}{2d\cos\phi} $$ $$ \tan\alpha = \tan\phi + \frac{gT^2}{2d\cos\phi} $$
Substitute back $x_B$ and $K$: $$ \tan\alpha = \frac{d\sin\phi + \frac{1}{2}gT^2}{d\cos\phi} $$ $$ \tan\alpha = \frac{d\sin\phi}{d\cos\phi} + \frac{gT^2}{2d\cos\phi} $$ $$ \tan\alpha = \tan\phi + \frac{gT^2}{2d\cos\phi} $$
Answer:
$$ \alpha = \tan^{-1} \left( \tan\phi + \frac{gT^2}{2d\cos\phi} \right) $$
$$ \alpha = \tan^{-1} \left( \tan\phi + \frac{gT^2}{2d\cos\phi} \right) $$