Question

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 m/s.

Answer 1

distance, s = 500m
acceleration, g = 10 m/s²
initial speed, u = 0
let the time taken for the stone to reach the pond = t₁

s = ut₁ + 1/2 gt₁²
⇒ 500 = 0 + (1/2)×10×t₁²
⇒ t₁² = (2×500)/10 = 1000/10 = 100
⇒ t₁ = √100 = 10s

let the time taken by sound to reach the top of tower = t₂
speed of sound = 340 m/s
t₂ = 500/340 = 1.47s

total time = t₁+t₂ = 10+1.47 = 11.47s
So the splash will be heard at the top after 11.47 seconds.

Answer 2

S = 500 m = distance to travel - for the stone

g = 10 m/sec/sec,        initial velocity of stone  u = 0

equation of the dynamics of stone :

     s = u t + 1/2 g t^2

       500 = 0  + 1/2 * 10 * t^2

  t =  10 sec

Immediately after touching the water, in a fraction of second, there is a sound wave produced on the surface of water - splashing of water.

time for sound to reach the top = distance to travel / speed = 

                 500 meter / 340 m/sec  = 25/17 sec

total time = 10 sec + 25/17 sec