Two shots are fired simultaneously from the bottom and top of a vertical tower $AB$ at angles $\gamma$ and $\beta$ with horizontal respectively. Both strike the same point $C$ on the ground at a distance $S$ from the bottom of the tower. What is the height of the tower?

The correct option is B $S(\tan \gamma -\tan \beta )$

$(S_{rel}{)}_x=0$
$\Rightarrow S=\left(u_1\cos \beta \right)t=\left(u_2\cos \gamma \right)t$
$\begin{array}{}\Rightarrow t=\frac{S}{u_1\cos \beta }=\frac{S}{u_2\cos \gamma }& \text{(1)}\end{array}$
Using 2nd equation of motion for vertical displacement of both the shots,
$\begin{array}{}h=-u_{1}t\sin \beta +\frac{1}{2}g{t}^2& \text{(2)}\end{array}$
$\begin{array}{}0=-u_{2}t\sin \gamma +\frac{1}{2}g{t}^2& \text{(3)}\end{array}$
Comparing equaion $\left(2\right)$ and $\left(3\right)$, we get
$\Rightarrow u_{1}t\sin \beta +h=u_{2}t\sin \gamma$
Substituting $t$ from equation $\left(1\right)$,
$\Rightarrow u_1\left(\frac{S}{u_1\cos \beta }\right)\sin \beta +h=u_2\left(\frac{S}{u_2\cos \gamma }\right)\sin \gamma$
$\Rightarrow h=S(\tan \gamma -\tan \beta )$