A stone is dropped from the top of a tower 500 m high into a pond of water at the base of the tower. When is the splash heard at the top? Given, g = 10 ms^-2 and speed of sound = 340 ms^-1

Given,

Height of tower (s) = 500 m

Velocity of sound (v) = 340 ms^-1

Acceleration due to gravity (g) = 10 ms^-2

Initial velocity of the stone (u) = 0

Time taken by the stone to fall to the tower base (t₁):

According to the second equation of motion,

s = ut₁ + $\dfrac{1}{2}$gt₁²

Substituting we get,

500 = (0 x t₁) + ($\dfrac{1}{2}$ x 10 x t₁²)

500 = $\dfrac{1}{2}$ x 10 x t₁²

1000 = 10 x t₁²

t₁² = $\dfrac{1000}{10}$ = 100

Hence, t₁ = $\sqrt{100}$ = 10 s

Time taken by sound to reach the top from the tower base (t₂)= ?

Using,

Speed = $\dfrac{\text{Distance}}{\text{Time}}$

or

Time = $\dfrac{\text{Distance}}{\text{Speed}}$

Substituting we get,

t₂ = $\dfrac{500}{340}$ = 1.47 s

t = t₁ + t₂

t = 10 + 1.47

t = 11.47 s

Hence, the splash will be heard at the top after 11.47 s.