Question

A cannon and a target are $5.10km$ apart and located at the same level. How soon will the shell launched with the initial velocity $240m/s$ reach the target in the absence of air drag?

Answer 1

Total time of motion is given by,
$\tau =\frac{2v_0\sin \alpha }{g}$ or $\sin \alpha =\frac{\tau g}{2v_0}=\frac{9.8\tau }{2\times 240}$ ...........................................................(1)
and horizontal range
$R=v_0\cos \alpha \tau$ or $\cos \alpha =\frac{R}{v_0\tau }=\frac{1500}{240\tau }=\frac{85}{4\tau }$ ...............................................(2)
From equations (1) and (2)
$\frac{{\left(9.8\right)}^2{\tau }^2}{{\left(480\right)}^2}+\frac{{\left(85\right)}^2}{{\left(4{\tau }^2\right)}^2}=1$
On simplifying ${\tau }^4-2400{\tau }^2+1083750=0$
Solving for ${\tau }^2$, we get,
${\tau }^2=\frac{2400\pm \sqrt{1425000}}{2}=\frac{2400\pm 1194}{2}$
Thus $\tau =42.39s=0.71min$ and
$\tau =24.55s=0.41min$ depending on the angle $\alpha$.