What will be the formula for range of the projectile if it is launched from the ground level but lands at some height about the ground? I have searched everywhere for the answer, but everyone has discussed the case of "launching from some height" but not "landing at some height". The picture of the scenario is attached below.
On
In a physical framework you may also perform some simplification, assuming that $200\,m$ are negligible with respect to $h_m$, as the picture seems to suggest. Removing the last step you have a range equal to $\frac{u^2}{g}\sin(2\theta)$, and assuming the last part of the trajectory to be approximately linear you have that the answer to your question is fairly close to $$ \frac{u^2}{g}\sin(2\theta)-f\cot\theta. $$ You can compare this with the exact expression found by Tavish:
$$\begin{eqnarray*} \frac{u^2 \sin2\theta + u\cos\theta\sqrt{4u^2\sin^2\theta -8gf}}{2g}&=&\frac{u^2\sin(2\theta)}{2g}\left(1+\sqrt{1-\frac{2gf}{u^2\sin^2\theta}}\right)\\&\approx&\frac{u^2\sin(2\theta)}{2g}\left(2-\frac{gf}{u^2\sin^2\theta}\right)\\&=&\frac{u^2}{g}\sin(2\theta)-f\cot\theta\end{eqnarray*}$$ and the approximation is accurate as soon as $gf\ll u^2\sin^2\theta$.
As usual, $$ x(t) = u\cos \theta t \\ y(t) = u\sin\theta t - \frac 12 gt^2 $$ Say the projectile lands a distance $f$ above the ground. Now, $$y (t) = f \\ \iff gt^2 -2u\sin\theta t +2f =0 \\ $$ The larger root of this quadratic is $$\frac{2u\sin\theta + \sqrt{4u^2\sin^2 \theta -8gf}}{2g}=\tau $$ Then, the range required will be the value of $x$ at this time, i.e. $$x(\tau) = \frac{u^2 \sin2\theta + u\cos\theta\sqrt{4u^2\sin^2\theta -8gf}}{2g}$$