Question

A man fires a bullet standing between two cliffs. first echo is heard after 3 s and the second echo is heard after 5 s. if the velocity of sound is 330 m/s, then please help me to calculate the distance between the cliffs.

Answer 1

First distance = d1
Second distance = d2
BTP,
d1 = (v×t)/2
(330×3)/2 = 495m

d2 = (v×t)/2
(330×5)/2 = 825m

.°. Total distance = d1+d2 = (495+825)m=1320m

Answer. The total distance between the cliffs is 1320m.

Answer 2

Given :

A  man fires a bullet standing between two cliffs

The time after which first echo heard = $t_1$ = 3 sec

The  time after which second echo heard = $t_5$ = 5 sec

The velocity of sound = v = 330 m/s

To Find :

The distance between the cliff

Solution :

Let The distance between the cliff = D meters

we know,

Distance = speed × time

Now,

For first echo

$D_1$ = 330 m/s × $t_1$

    = 330 m/s × 3 sec

    = 990 meters

Again

For second echo

$D_2$ = 330 m/s × $t_2$

    = 330 m/s × 5 sec

    = 1650 meters

So, Total Distance cover by echo = $D_1+D_2$

Or, Distance = 990 meters + 1650 meters

                       = 2640 meters

Hence, The Distance between the cliff us 2640 meters  . Answer