A gun can fire shells with maximum speed $\boldsymbol{v}_0$ and the maximum horizontal range that ca

Question

A gun can fire shells with maximum speed $\boldsymbol{v}_0$ and the maximum horizontal range that can be achieved is $R=\frac{v_0^2}{g}$. If a target farther away by distance $\Delta x$ (beyond $R$ ) has to be hit with the same gun (see figure), then it could be achieved by raising the gun to a height at least

Professor PC Thomas Classes · Answer

(a) : Maximum range, $R=\frac{v_0^2}{g}$ Here, $\sin 2 	heta=1$ \ $ \begin{array}{ll} \Rightarrow & 2 	heta=90^{\circ} \ \Rightarrow & 	heta=45^{\circ} \end{array} $ \ Consider the motion of a shell after point $P$. \ For horizontal direction, \  $ \Delta x=\left(v_0 \cos 	hetaight) t \Rightarrow t=\frac{\Delta x}{v_0 \cos 	heta}  $ \ For vertical direction, $h=\left(v_0 \sin 	hetaight) t+\frac{1}{2} g t^2$ 
$\begin{aligned} & =\left(v_0 \sin 	hetaight)\left(\frac{\Delta x}{v_0 \cos 	heta}ight)+\frac{1}{2} g\left(\frac{\Delta x}{v_0 \cos 	heta}ight)^2 \ & \Rightarrow h=\Delta x\left[	an 	heta+\frac{\Delta x}{2\left(\frac{v_0^2}{g}ight) \cos ^2 	heta}ight]\end{aligned}$
$\Rightarrow h=\Delta x\left[	an 45^{\circ}+\frac{\Delta x}{2 R \cos ^2 45^{\circ}}ight]$
$\Rightarrow h=\Delta x\left[1+\frac{\Delta x}{R}ight]$