Q. 24.4( 12 Votes )

# A stone dropped from the top of a tower of height 300 m high splashes into the water of a pond near the base of the tower. When is the splash heard at the top given that the speed of sound in air is 340 m s^{–1}? (g = 9.8 m s^{–2})

Answer :

Given:

Height of the tower, h = 300 m

Initial velocity of the stone, u = 0 ms^{-1}

Acceleration due to gravity, g = 9.8 ms^{-1}

Speed of the sound in air = 340 ms^{-1}

The time taken by the stone to hit the pond can be found by the second equation of motion,

s = ut + 1/2 gt^{2} …(1)

Where,

s = distance, in this case height

u = initial velocity

g = acceleration due to gravity

t = time

Putting the values in equation (1), we get

s = ut + 1/2 gt^{2}

t = (2s/g)^{1/2}

t = [2× (300m/9.8 ms^{-2})]^{1/2}

t = 7.82 s

Time taken by the sound of splash to the top of the tower can also be given by,

t_{1} = h/v …(2)

Where,

h = height of the tower

v = velocity of sound (340 m/s)

t_{1} = 300m/340ms^{-1}

t_{1} = 0.88 s

The total time after which splash is heard is equal to sum of the time taken by the stone to reach the water and time taken by sound to reach the tower.

i.e. T = t + t_{1}

T = 7.82s + 0.88 s

T = 8.7 s

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