Question

A stone is dropped from the top of a tower 500m high into a ponds of water at the base of tower. When is the splash heard at the top? Given g=10m/s and speed of sound=300m/s

Answer 1

H = 500m
g = 10m/s
v = 300m/s

s = ut+1/2at²
500 = 1/2×10×t²
1000 = 10t²
t² = 1000/10
t = 10s ____(i)

t = d/v = 500/300 = 1.6s ____(ii)

Time = (i) + (ii)
= 10+1.6
= 11.6s

Answer 2

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• Height of the tower, s = 500 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:

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

$\rm 500 = 0 \times t_1 + 1/2 \times 10 \times t_1^2$

$\rm t_1^2 = 100$

$\rm t_1 = 10s$

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, $\rm t= t_1 + t_2 = 10 + 1.47 = 11.47 s.$