A gun can fire shells with maximum speed $v_0$ and the maximum horizontal range that can be achieved is $R_{max} = \frac {v_0^2}{g}$.  If a target farther away by distance $\Delta x$ (beyond $R$) has to be hit with the same gun, show that it could be achieved by raising the gun to a height at least $h = \Delta x\,\left[ {1 + \frac{{\Delta x}}{R}} \right]$.

This problem can be approached in two different ways.

$(i)$ Refer to the diagram target $\mathrm{T}$ is at horizontal distance $x=\mathrm{R}+\Delta x$ and between point projection $y=-h$.

$(ii)$ From point $\mathrm{P}$ in the diagram projection at speed $v_{0}$ at an angle $\theta$ below horizontal with height $h$ and horizontal range $\Delta x \mathrm{~A}$ )

Applying method $(i)$, Maximum horizontal range, $\mathrm{R}=\frac{v_{0}^{2}}{g}$, for $\theta=45^{\circ}$

Let the gun be raised through a height $h$ from the ground so that it can hit the target. Let vertically downward direction is taken as positive.

Horizontal component of initial velocity $=v_{0} \cos \theta$

Vertical component of initial velocity $=-v_{0} \sin \theta$

Taking motion in vertical direction, $h=\left(-v_{0} \sin \theta\right) t+\frac{1}{2} g t^{2}$

Taking motion in horizontal direction, $(\mathrm{R}+\Delta x)=v_{0} \cos \theta \times t$

$\therefore t=\frac{(\mathrm{R}+\Delta x)}{v_{0} \cos \theta}$

Substituting value of $t$ in equ. (ii) we get

$h$$=-v_{0} \sin \theta \times\left(\frac{ R _{\max }+\Delta x}{v_{0} \cos \theta}\right)+\frac{1}{2} g\left(\frac{ R _{\max }+\Delta x}{v_{0} \cos \theta}\right)^{2}$

$h$$=-\tan \theta\left( R _{\max }+\Delta x\right)+\frac{1}{2} \frac{g\left( R _{\max }+\Delta x\right)^{2}}{v_{0}^{2} \cos \theta}$