Hint: Time is taken to travel horizontal and vertical distances will be the same.

Step 1: Find the maximum range.
$\mathrm{R} = \frac{{\mathrm{v}}_0^2}{\mathrm{g}},\text{ for }\mathrm{\theta } = {45}^{\circ } ...\left(\mathrm{i}\right)$

Step 2: Apply equation in the horizontal direction

The horizontal component of initial velocity =$v_0\cos \theta$

$\begin{array}{l}(\mathrm{R}+\mathrm{\Delta x})={\mathrm{v}}_0\mathrm{cos\theta }\times \mathrm{t}\\ \Rightarrow \mathrm{t}=\frac{(\mathrm{R}+\mathrm{\Delta x})}{{\mathrm{v}}_0\mathrm{cos\theta }}\end{array}$                                      ...(ii)

Step 3: Put the value of t in from equation (ii) for vertical motion and calculate h

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

The vertical component of initial velocity $=-v_0\sin \theta$
Taking motion in the vertical direction, $h=(-v_0\sin \theta )t+\frac{1}{2}g{t}^2$                 ...(iii)

Substituting the value of t in Eq. (iii), we get
$\begin{array}{l}\mathrm{h}=(-{\mathrm{v}}_0\mathrm{sin\theta })\times \left(\frac{\mathrm{R}+\mathrm{\Delta x}}{{\mathrm{v}}_0\mathrm{cos\theta }}\right)+\frac{1}{2}\mathrm{g}(\frac{\mathrm{R}+\mathrm{\Delta x}}{{\mathrm{v}}_0\mathrm{cos\theta }}{)}^2\\ \mathrm{h}=-(\mathrm{R}+\mathrm{\Delta x})\mathrm{tan\theta }+\frac{1}{2}\mathrm{g}\frac{(\mathrm{R}+\mathrm{\Delta x}{)}^2}{{\mathrm{v}}_0^2{\cos }^2\mathrm{\theta }}\end{array}$
As the angle of projection is $\mathrm{\theta } = 45°$, therefore

$\begin{array}{l}\mathrm{h}=-(\mathrm{R}+\mathrm{\Delta x})+\tan {45}^{\circ }+\frac{1}{2}\mathrm{g}\frac{(\mathrm{R}+\mathrm{\Delta x}{)}^2}{{\mathrm{v}}_0^2{\cos }^2{45}^{\circ }}\\ \mathrm{h}=-(\mathrm{R}+\mathrm{\Delta x})\times 1+\frac{1}{2}\mathrm{g}\frac{(\mathrm{R}+\mathrm{\Delta x}{)}^2}{{\mathrm{v}}_0^2(1/2)}\end{array}$
$\left(\because \tan {45}^{\circ }=1\text{ and }\cos {45}^{\circ }=\frac{1}{\sqrt{2}}\right)$
$\begin{array}{l}\mathrm{h}=-(\mathrm{R}+\mathrm{\Delta x})+\frac{(\mathrm{R}+\mathrm{\Delta x}{)}^2}{\mathrm{R}} \left[\mathrm{using} \mathrm{EQ}. \left(\mathrm{i}\right) \mathrm{R} = {\mathrm{v}}_0^2/\mathrm{g}\right]\\ =-(\mathrm{R}+\mathrm{\Delta x})+\frac{1}{\mathrm{R}}({\mathrm{R}}^2+{\mathrm{\Delta x}}^2+2\mathrm{R\Delta x})\\ =-\mathrm{R}-\mathrm{\Delta x}+(\mathrm{R}+\frac{{\mathrm{\Delta x}}^2}{\mathrm{R}}+2\mathrm{\Delta x})\\ =\mathrm{\Delta x}+\frac{{\mathrm{\Delta x}}^2}{\mathrm{R}}\\ \mathrm{h}=\mathrm{\Delta x}(1+\frac{\mathrm{\Delta x}}{\mathrm{R}})\end{array}$