Let the distance of cliff from the initial position of man be ' $\mathit{d}$ ' metres and speed of sound in air be ' $\mathrm{V}$ ' $\mathrm{m}∕\mathrm{}\mathit{s}$.
For the first echo,
${\mathit{t}}_1=3.5\mathrm{}\mathit{s}$
$\mathrm{V}=\frac{2\mathit{d}}{{\mathit{t}}_1}$
$\therefore \mathrm{V}=\frac{2\mathit{d}}{3\cdot 5}$ ......(i)
On moving closer to the cliff by $84\mathrm{m}$, the distance of cliff from the new position becomes $\left(\mathit{d}-84\right)\mathrm{m}$, then
for second echo, ${\mathit{t}}_2=3\mathrm{}\mathit{s}$.
$\because \mathrm{V}=\frac{2\left(\mathit{d}-84\right)}{{\mathit{t}}_2}$
$\therefore \mathrm{V}=\frac{2\left(\mathit{d}-84\right)}{3}$ .......(ii)
From (i) and (ii), we get
$\frac{2\left(\mathit{d}-84\right)}{3}=\frac{2\mathit{d}}{3\cdot 5}$
$\Rightarrow 3\cdot 5\left(\mathit{d}-84\right)=3\mathit{d}$
$\Rightarrow 3\cdot 5\mathit{d}-294=3\mathit{d}$
$\Rightarrow 3\cdot 5\mathit{d}-3\mathit{d}=294$
$\Rightarrow 0.5\mathit{d}=294$
$\Rightarrow \mathit{d}=\frac{294}{0.5}=588\mathrm{m}.$