Question

A stone is dropped from the top of a cliff. The splash it makes when striking the water below is heard 4.1s later. The speed of sound in air is 343 m/s. How high is the cliff?

Answer 1

Let t₁ be the time taken by the rock to touch the water and

t₂ be the time taken by the sound to reach us.

Given that t₁ + t₂ = 4.1 s    -------------(1)

Velocity of sound v = 343 m/s

Let S be the distance tavelled by the rock.
          t₂ = S/v = S/343 ........... (2)
from eq of motion, S = (1/2)g(t₁)²    (initial speed u =0)
         t₂ = [(1/2)(9.8 m/s²)(t₁)²]/(343 m/s)
     = 0.014286(t₁)² ..........(3)
substitute the eq (3) in eq (1), we get
     0.014286(t₁)² + t₁ - 4.1 = 0
On solving the above quadratic equation we get
                      t₁ = 3.884 s

Hence distance covered by the rock is

                      S = (1/2)(9.8 m/s²)(3.884)²

                         = 73.92 m

Answer 2