Question

If an observer standing between two cliffs receives echo at 1.5 s and 4 s after clapping,find the distance between the cliffs if the velocity of sound is 320 m/s

Answer 1

Case 1
t = 1.5s
V = 320
d = Vt/2
= 320x1.5/2
=240m

Case 2
t = 4s
V = 320
d = Vt/2
= 320x4/2
= 640m

Distance between the cliffs = 240+640 = 880m

Answer 2

Distance between the cliffs = 880 m

Explanation:

Clapping Echo - first instance

time t = 1.5 seconds

Velocity of sound V = 320 m/s

distance = velocity * time / 2 = Vt / 2

= 320 x 1.5 / 2

= 240 metres

Clapping Echo - second instance

t = 4s

V = 320

d = Vt / 2

= 320 x 4 / 2

= 640 metres

So the distance between the cliffs would be the sum of the distance of both these echoes = 240 + 640 = 880 metres