Question

A shot is fired at angle ${30}^o$ with horizontal from tower of height $182.88m$. Projection velocity is $60.96m/s$. Find where from the foot of tower it strkes the ground.

Answer 1

Height from which shot id fired, $h=182.88m$

Vertical component of the initial velocity, $u_y=60.96\sin 30=30.48m/s$

Using,

$s=ut+\frac{1}{2}a{t}^2$ for vertical displacement

$182.88=-30.48t+\frac{1}{2}\times 9.8t^2$

$t=9.96sec$

In this time the shot moves forward with its horizontal velocity component ,

$u_x=60.96\cos 30=52.56m/s$

So, horizontal displacement of the shot is

$X=52.56\times 9.96=523.5m$