Question

A stone is dropped from the top
and it hits the ground at t = 3.5 s. If the veloc-
ity of sound in air is 300 m s -1, find the time
taken to hear the sound by a person on the top of
the tower, from the instant the body is dropped.
(g = 10 m s-2)​

Answer 1

Answer:

3.7 sec

Explanation:

Using S = ut + ½at² ; here,

As body is dropped,

u = 0, a = g(gravity), time = 3.5

Therefore,

S = (0)t + ½(10)(3.5)² = 61.25 m.

Means, height of that top is 61.25 m.

As the sound is produced at the bottom, it must cover the same distance of 61.25 m to reach to the man.

=> speed = distance/time

=> 300 = 61.25/time

=> time ≈ 0.20 sec

Hence, total time to sound to be produced is 3.5 + 0.20 = 3.7 sec

Answer 2

Answer:

Given :

• A stone is dropped from the top and it hits the ground at t = 3.5 s.

• If the velocity of sound in air is 300 m s -1,

To Find :

• find the timetaken to hear the sound by a person on the top of the tower, from the instant the body is dropped.

Solution :

$: \implies \: \: \: \: \: \boxed{ \sf \: S \: = ut \: + \frac{1}{2} {at}^{2} }$

Substitute all values :

$: \implies \: \: \: \: \: \sf \: S \: = 0 \times t \: + \frac{1}{2} \times {10 \times 3.5 \: }^{2} \\ \\ \\ : \implies \: \: \: \: \: \sf \: S \: = \frac{\cancel{10} \times \: 12.25 }{ \cancel{2} } \\ \\ \\ : \implies \: \: \: \: \: \sf \: S \: = 61.25$

$: \implies \: \: \: \: \: \boxed{ \sf \:speed = \frac{distance}{time}}$

Substitute all values :

$: \implies \: \: \: \: \:\sf \: \: 300 = \frac{61.25}{time}\\ \\ \\ : \implies \: \: \: \: \:\sf \: \: \: time = \frac{61.25}{300} \\ \\ \\ : \implies \: \: \: \: \:\sf \: \: \: time = 0.20 \: sec$

Total Time :

$\implies \: \: \: \: \:\sf \: \: \: \: 3.5 + 0.20 \\ \\ \implies \: \: \: \: \:\sf \: \: \: \: 3.70$

Hence, Total time to sound to be produced is 3.7 sec