The position of the catapult along the wall isn't important because it always shoots perpendicular to the wall. In the catapult, the trajectory of the projectile is defined by
\(\begin{aligned}{c}x = {v_0}\cos \alpha \cdot t\\y = {v_0}\sin \alpha \cdot t - \frac{1}{2}g{t^2}\end{aligned}\)
Since the velocity at which the projectile is launched is \(80\frac{m}{s}\) and taking,
\(g = 9.8\frac{m}{{{s^2}}}\) we get that
\(\begin{aligned}{c}x = 80\cos \alpha \cdot t\\y = 80\sin \alpha \cdot t - 4.9{t^2}\end{aligned}\)
The first condition is:
\(d > 100\)
Where \(d\)is the bullet's horizontal distance, ensuring that the projectile hits the wall.
The second condition is:
\(y\left( {{t_0}} \right) > 15\)
Where \({t_0}\)is the value of \(t\)for which \(x(t) = 100\). This assures that when the missile reaches the wall, it will cross it.
Because the horizontal distance is the same as the vertical distance
\(\begin{aligned}{c}d = \frac{{v_0^2}}{g} \cdot \sin 2\alpha \\ = \frac{{{{80}^2}}}{{9.8}} \cdot \sin 2\alpha \end{aligned}\)
The first condition is equivalent to
\(\begin{aligned}{c}100 < \frac{{{{80}^2}}}{{9.8}} \cdot \sin 2\alpha \Leftrightarrow \\980 < 6400 \cdot \sin 2\alpha \Leftrightarrow \\\frac{{49}}{{320}} < \sin 2\alpha \end{aligned}\)
Since \({0^^\circ } < \alpha < {90^^\circ }\) we have that \({0^^\circ } < 2\alpha < {180^^\circ }\) so the solution to the inequality above is
\(\begin{aligned}{c}{\sin ^{ - 1}}\left( {\frac{{49}}{{320}}} \right) < 2\alpha < {180^^\circ } - {\sin ^{ - 1}}\left( {\frac{{49}}{{320}}} \right) \Leftrightarrow \\{8.8^^\circ } < 2\alpha < {180^^\circ } - {8.8^^\circ } \Leftrightarrow \\{8.8^^\circ } < 2\alpha < {171.2^^\circ } \Leftrightarrow {4.4^^\circ } < \alpha < {85.6^^\circ } \ldots (1)\end{aligned}\)
As a result, we've discovered the first requirement that must be met. We shall solve the equation assuming (1) is true\(x(t) = 100\) for \(t\).
The solution is:
\(\begin{aligned}{c}x(t) = 100 \Leftrightarrow \\80\cos \alpha \cdot t = 100 \Leftrightarrow \\{t_0} = \frac{{100}}{{80\cos \alpha }}\\ = \frac{5}{{4\cos \alpha }}\end{aligned}\)
Therefore, the second condition is identical to
\(\begin{aligned}{*{20}{r}}{y\left( {\frac{5}{{4\cos \alpha }}} \right) > 15}\\{80\sin \alpha \times \frac{5}{{4\cos \alpha }} - 4.9 \times {{\left( {\frac{5}{{4\cos \alpha }}} \right)}^2} > 15}\\{100\tan \alpha - 4.9 \times \frac{{25}}{{16}} \times \frac{1}{{{{\cos }^2}\alpha }} > 15}\\{100\tan \alpha - 7.65625 \times \left( {1 + {{\tan }^2}\alpha } \right) > 15}\\{7.65625 \times \left( {1 + {{\tan }^2}\alpha } \right) - 100\tan \alpha < - 15}\\{7.65625{{\tan }^2}\alpha - 100{{\tan }^2}\alpha + 22.65625 < < 00}\end{aligned}\)
The equivalent quadratic equation for \(\tan \alpha \) have the following solutions: \(\tan \alpha = 0.230635{\rm{ }}and{\rm{ }}\tan \alpha = 12.8306\)
And for \(\tan \alpha \), the solution to the inequality is:
\(0.230635 < \tan \alpha < 12.8306\)
The solution for \(\alpha \)is:
\({\tan ^{ - 1}}(0.230635) < \alpha < {\tan ^{ - 1}}(12.8306) \Leftrightarrow {12.99^^\circ } < \alpha < {85.54^^\circ }.\)
We can deduce from the intersection of the intervals found in \(({\bf{1}})\) and \(({\bf{2}})\) that the range of angles we should set the catapult to (so far) is:
\({12.99^^\circ } < \alpha < {85.54^^\circ }\)
