$A$ gun can fire shells with maximum speed $v_0$ 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,show that it could be achieved by raising the gun to a height at least $h = \Delta x \left[ 1 + \frac{\Delta x}{R} \right]$.

(A) The equation of the trajectory of a projectile launched from a height $h$ with speed $v_0$ at an angle $\theta$ with the horizontal is given by:
$y = x \tan \theta - \frac{gx^2}{2v_0^2 \cos^2 \theta}$
Let the target be at a horizontal distance $x = R + \Delta x$ and vertical position $y = -h$ (taking the launch point as the origin).
$-h = (R + \Delta x) \tan \theta - \frac{g(R + \Delta x)^2}{2v_0^2 \cos^2 \theta}$
Using $R = \frac{v_0^2}{g}$,we have $\frac{g}{v_0^2} = \frac{1}{R}$. Substituting this:
$-h = (R + \Delta x) \tan \theta - \frac{(R + \Delta x)^2}{2R \cos^2 \theta}$
$-h = (R + \Delta x) \tan \theta - \frac{(R + \Delta x)^2}{2R} (1 + \tan^2 \theta)$
Rearranging as a quadratic in $\tan \theta$:
$\frac{(R + \Delta x)^2}{2R} \tan^2 \theta - (R + \Delta x) \tan \theta + \left[ \frac{(R + \Delta x)^2}{2R} - h \right] = 0$
For a real solution for $\tan \theta$,the discriminant $D \ge 0$:
$D = (R + \Delta x)^2 - 4 \left[ \frac{(R + \Delta x)^2}{2R} \right] \left[ \frac{(R + \Delta x)^2}{2R} - h \right] \ge 0$
$(R + \Delta x)^2 - \frac{(R + \Delta x)^4}{R^2} + \frac{2h(R + \Delta x)^2}{R} \ge 0$
Dividing by $(R + \Delta x)^2$:
$1 - \frac{(R + \Delta x)^2}{R^2} + \frac{2h}{R} \ge 0$
$\frac{2h}{R} \ge \frac{(R + \Delta x)^2 - R^2}{R^2} = \frac{R^2 + 2R\Delta x + \Delta x^2 - R^2}{R^2} = \frac{2R\Delta x + \Delta x^2}{R^2}$
$h \ge \frac{2R\Delta x + \Delta x^2}{2R} = \Delta x + \frac{\Delta x^2}{2R} = \Delta x \left[ 1 + \frac{\Delta x}{2R} \right]$
Note: The expression in the question $h = \Delta x [1 + \frac{\Delta x}{R}]$ is a common approximation or specific case result; the derivation shows the minimum height required is $h = \Delta x (1 + \frac{\Delta x}{2R})$.