Question

A shot fired from a gun with a speed of $V$ at an angle $\theta$ strikes an object at a point P on the horizontal plane through the point of projection. lf the object at P starts to move away from the gun with uniform acceleration a when the gun is fired, then the elevation is to be changed to $\varphi$ in order to strike the moving object.

lf $\frac{g}{a}=\left(\frac{1-cosn\varphi }{sinn\varphi -sinn\theta }\right)$, find $n$.

Answer 1

Initially, horizontal range $R=\frac{v^2\sin 2\theta }{g}$

Time of flight, $T=\frac{2v\sin \varphi }{g}$

Then distance moved by P in time T is given by,
$S=\frac{1}{2}a{\left(\frac{2v\sin \varphi }{g}\right)}^2$

$\therefore$ changed range = R+S

$\Rightarrow \frac{v^2\sin 2\varphi }{g}=\frac{v^2\sin 2\theta }{g}+\frac{2a{v}^2{\sin }^2\varphi }{g^2}$

$\Rightarrow g\sin 2\varphi =g\sin 2\theta +2asi{n}^2\varphi$

RHS $=g\sin 2\theta +2a\left(\frac{1-\cos 2\varphi }{2}\right)$
$=g\sin 2\theta +a-acos2\varphi$

$\Rightarrow g\sin 2\varphi -g\sin 2\theta =a-acos2\varphi$

$\Rightarrow g(\sin 2\varphi -\sin 2\theta )=a(1-\cos 2\varphi )$

$\Rightarrow \frac{g}{a}=\frac{1-\cos 2\varphi }{\sin 2\varphi -\sin 2\theta }$

$\Rightarrow n=2$