Look at the modified diagram above.
The dark lines and the arrows show the directions in which sound travels from P to Q.
The 1st sound, which is heard by the person at Q, travels along the straight line PQ.
The second sound travels along the path POQ. Here the sound gets reflected at point O on the cliff.
The thin line OR is the Normal of the Reflection.
As the path POQ is longer than PQ, the 2nd sound reaches the person's ear 2 seconds later than the time when the 1st sound reached.
That is why the person heard the sound twice.
The driver heard an echo because the sound travelled along the perpendicular distance between the cliff and him, got relfected from the cliff, moved backwards along the same direction and reached his ear.
Thus the sound has travelled twice the distance 680m i.e. 1360m.
Since speed(v)=distancetime the speed of sound is v=6804=170m/s
PO2=PR2+OR2 ------------------------------------------(1)
where OR=680
Since PR=RQ,PQ=2PR and POQ=2PO
Time taken by the 1st sound to reach Q =PQv=2PRv
Time taken by the 2nd sound to reach Q =POQv=2POv
Since the time interval between hearing the 1st sound and the 2nd is 2 seconds, 2POv−2PRv=2
∴PO−PR=v=170
∴PO=170+PR
Substituting in (1) we get
(170+PR)2=PR2+6802
∴1702+(2×170×PR)+PR2=PR2+6802
∴1702+340PR=6802
∴PR=6802−1702340=1275
∴PQ=2PR=2550metres
The person was 2550 metres away from the truck