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 $21m/\sec$. Find the distance of point from the gun at which shell will fall.

Answer 1

Answer: 30.00

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