A ship travelling towards a cliff receives the echo of its whistle after 3.5 seconds. A short while later, it receives the echo after 2.5 seconds. If the speed of sound in air under the prevalling condition is 250ms ^-1, how much closer is the ship to the cliff?

fryoscience19

9 years ago

Here is an explanation:

Using the formula

V=2d/t

V=250ms

t1=3.5s

t2=2.5s

therefore

250=2d/3.5

cross multiplie

250x3.5= 87.5ms/2

d=437.5ms

AND

250=2d/2.5

same method

250X2.5=625ms/2

d=312.5ms

now d= 437.5-312.5

d=125ms

REF: Check new school physic page 331

DominicAfam

1 year ago

T1 -T2
T=1

V=2d/t
250=2d/1
2d=250
Divide both side by 2
d=125m/s

Goodluck.

Nkiru555

1 year ago

sharp bro

Clintonchuks998

2 years ago

thanks bro 🥹