Question

A shell is fired from a gun from the bottom of a hill along its slope. The slope of the hill is $\alpha=30^{\circ}$, and the angle of the barrel to the horizontal $\beta=60^{\circ}$. The initial velocity $v$ of the shell is $21\, m / \sec$. Find the distance of point from the gun at which shell will fall.

Answer 1

Take $x$ and $y$ axis as shown in figure. Let shell falls at point A on the incline along $y$-direction
$u_{y}=21 \sin 30^{\circ}=\frac{21}{2}$
$a_{y}=-g \cos 30^{\circ}=-5 \sqrt{3}$
$s_{y}=0$
$s_{y}=u_{y} t+\frac{1}{2} a_{y} t^{2}$
$0 =\frac{21}{2} t+\frac{1}{2}(-5 \sqrt{3}) t^{2}$
$t=\frac{21}{5 \sqrt{3}}$
$s_{x} =u_{x} t+\frac{1}{2} a_{x} t^{2}$
$=21 \cos 30^{\circ} \times \frac{21}{5 \sqrt{3}}+\frac{1}{2}\left(-g \sin 30^{\circ}\right)\left(\frac{21}{5 \sqrt{3}}\right)^{2}$
$=\frac{(21)^{2}}{10}-\frac{(21)^{2} \times 10}{300}=44.1-14.7$
$=29.4 \approx 30\, m$