Question

A stone is dropped from the top of a tower 125m high into a pond of water at the base of tower. When is splash heard at the top?

Answer 1

At the time the stone is dropped the stone is at rest. So,

u=0

and, h= 125m (given)

By using 2nd equation if motion for freely falling bodies:

$h = ut + 1 \div 2 \: gt {}^{2}$

For, u=0

Then,

$h = 1 \div 2 \: gt {}^{2}$

So,

$t = \sqrt{2h \div g}$

so,

$t = \sqrt{2 \times 125 \div 10}$

t= 5s

Answer 2

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Height of the tower, s = 250 m

Velocity of sound, v = 340 m sˆ’1

Acceleration due to gravity, g = 10 m sˆ’2

Initial velocity of the stone, u = 0 (since the stone is initially at rest)

Time taken by the stone to fall to the base of the tower, t1

According to the second equation of motion:

$\rm S = ut_1 + 1/2 g t_1^2$

$\rm 240 = 0 \times t_1 + 1/2 \times 10 \times t_1^2$

$\rm t_1^2 = 25$

$\rm t_1 = 5s$

Now, time taken by the sound to reach the top from the base of the tower,

$\rm t_2= 500/340 = 1.47 s$

Therefore, the splash is heard at the top after time, t

Where, $\rm t= t_1 + t_2 = 5 + 1.47 = 6.47 s.$