Question

A shell is fired from point O on the level ground with velocity $50m/s$ at angle ${53}^{\circ }$ . A hill of uniform slope ${37}^{\circ }$ starts from point A that is $100m$ away from the point O as shown in the figure. Calculate the time of flight (in seconds).

Answer 1

By using $y=xtan\theta =(1-\frac{x}{R})$ where $R=\frac{2u^{2}sin\theta cos\theta }{g}$ = 240m

$\frac{3}{4}x=(100+x)tan{53}^{\circ }[1-\frac{x+100}{240}]\Rightarrow x^2+95x-14000=0\Rightarrow x=\frac{-95\pm \sqrt{(95{)}^2+4(14000)}}{2}$ = 80

Required distance AB = $\frac{x}{cos{37}^{\circ }}=\frac{80}{4/5}$ = 100m

Time of fight = $\frac{100+x}{50cos{53}^{\circ }}=\frac{100+80}{50\left(3/5\right)}$ = 6s