Ans: (C)
Hint \(: S+n=\frac{1}{2} f(t+m)^{2}\) and \(S=\frac{1}{2} h t^{2}, V=h t\)
$$
\begin{aligned}
&\therefore \frac{1}{2} h t^{2}+n=\frac{1}{2} f(t+m)^{2}-(1) \\
&\text { Also } V=0+h t=0+f(t+m) \quad \Rightarrow t+m=\frac{h t}{f}
\end{aligned}
$$
From equation (1),
$$
\frac{1}{2} h t^{2}+n=\frac{1}{2} f\left(\frac{h t}{f}\right)^{2} \Rightarrow t^{2}=\frac{2 n f}{h(h-f)}
$$
Also,
$$
\begin{aligned}
&h t=f(t+m) \quad \Rightarrow t^{2}=\frac{m^{2} f^{2}}{(h-f)^{2}} \\
&\therefore \frac{2 n f}{h(h-f)}=\frac{m^{2} f^{2}}{(h-f)^{2}} \quad \Rightarrow 2 n=\frac{m^{2} f h}{h-f} \quad \Rightarrow n(h-f)=\frac{1}{2} f h m^{2}
\end{aligned}
$$