Let the distance of cliff from the initial position of man be ' $d$ ' metres and speed of sound in air be ' $V$ ' $\frac{m}{s}$. For the first echo, $t_1=3.5 \text{ s}$ $V=\frac{2d}{t_1}$ $\therefore V=\frac{2d}{3 \cdot 5}$ ......(i) On moving closer to the cliff by $84 \text{ m}$, the distance of cliff from the new position becomes $(d-84) \text{ m}$, then for second echo, $t_2=3 \text{ s}$. $\because V=\frac{2(d-84)}{t_2}$ $\therefore V=\frac{2(d-84)}{3}$ .......(ii) From (i) and (ii), we get $\frac{2(d-84)}{3}=\frac{2d}{3 \cdot 5}$ $\Rightarrow 3 \cdot 5 (d-84)=3d$ $\Rightarrow 3 \cdot 5 d-294=3d$ $\Rightarrow 3 \cdot 5 d-3d=294$ $\Rightarrow 0.5 d=294$ $\Rightarrow d=\frac{294}{0.5}=588 \text{ m}$