Question

A stone is dropped from the top of the tower 500m high into the pond of water at the base of tower. When is the splash heard at the top of the tower. Given, g=10m/s2 and speed of sound is=340m/s

Answer 1

Answer:

11.47 seconds

Explanation:

The time taken for the stone to reach the ground:

s = ut + 1/2 at²

500 = 0(t) + 1/2 x 10 x t²

500 = 5t²

Thus, t² = 100

∴ t = 10 seconds

Thus, after 10 seconds, the stone will splash into the water

For the sound to reach the observer, it would have to travel 500m upwards again from the pond.

The time taken for the sound to reach the observer from the pond:

500 ÷ 340 = 1.47 seconds

The time after which the observer hears the splash is = Time taken for the stone to reach the ground + Time taken for the sound to reach the observer

Thus,

Time after which splash is heard = 10+1.47

∴ Time after which splash is heard = 11.47 seconds

Hence, the splash is heard after 11.47 seconds

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