Answer to Question #190221 in Physics for Christian

Question

A cannon fires a shell upward from the top of a cliff 30m high. If 5s later it hits target on the level ground which is 2km away, find its muzzle velocity and the angle of inclination and find the maximum height reached with respect to the ground.

Answer 1

$\vec r=\vec v_0t+\vec gt^2/2$

$v_0t\sin(\alpha+\beta )=\frac{gt^2}{2}\cos\beta$

$\cos \beta=\frac{2000}{\sqrt{30^2+2000^2}}=0.999887\to \beta=0.8613°$

$v_0\cos\alpha\cdot t=2000\to v_0\cos \alpha=2000/5=400\ (m/s)$

$v_0t\sin(\alpha+0.8613 )=\frac{9.81\cdot 5^2}{2}\cos0.8613$

So, we have

$5v_0\sin(\alpha+0.8613 )=122.61\to v_0\sin(\alpha+0.8613 )=24.52$

$v_0\sin(\alpha+0.8613 )=24.52$

$v_0\cos \alpha=400$

$\frac{\sin(\alpha+0.8613 )}{\cos\alpha}=0.0613\to\alpha=2.65°$ . Answer

$v_0=400/\cos2.65°=400.43\ (m/s)$ . Answer

$h=v_0^2\sin^22.65°/(2g)=400.43^2\cdot\sin^22.65°/(2\cdot9.81)=17.45\ (m)$

$H=30+17.45=47.45\ (m)$ . Answer