Question

A stone is dropped from the top of a tower 750 m high into a pond of water at the base of the tower. When is the splash heard at the top?

Answer 1

Answer:

Given - h=500m,g10m/s

2

,v=340m/s ,

by , h=ut+(1/2)gt

2

,

500=0+(1/2)10t

2

, (initial velocity , u=0) ,

or t

2

=1000/10=100 ,

or t=10s ,

it is the time taken by stone to reach the water level , after that a sound is produced due to strike of stone on water , and sound travels upwards .Let t' be the time taken by sound to reach the base of tower ,

then , t

′

=h/v=500/340=1.47s ,

therefore time taken by splash to hear at the top ,

T=t+t

′

=10+1.47=11.47s

Answer 2

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Height of the tower, s = 750 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:

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

$\rm 750 = 0 \times t_1 + 1/2 \times 10 \times t_1^2$

$\rm t_1^2 = 150$

$\rm t_1 = 12.24s$

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, $\rm t= t_1 + t_2 = 12.24 + 1.47 = 13.71 s.$