Question

A stone is dropped from the top of a cliff. The splash it makes when striking the water below is heard 3.0s later. How high is the cliff?

Answer 1

Answer:

$40.7$ m

Explanation:

Consider the motion of the stone in vertical direction:

$v_{oy}$ = initial velocity = 0 m/s

$a_{y}$ = acceleration due to gravity = 9.8 m/s²

$t$ = time taken  to hit the water below

$h$ = vertical height of the cliff

Using the kinematics equation

$h = v_{oy} t + (0.5) a_{y} t^{2}$

$h = (0) t + (0.5) (9.8) t^{2}$

$h = 4.9 t^{2}$                                            

$t = \sqrt{\frac{h}{4.9}}$                           eq-1

consider the motion of sound from water in upward direction

$v_{s}$ = Speed of sound = 343 m/s

$t'$ = time of travel for sound

Using the equation

$h = v_{s} t'$    

$h = 343 t'$                                            

$t' = \frac{h}{343}$                                 eq-2

we have been given that

$t + t' = 3$

Using eq-1 and eq-2

$\sqrt{\frac{h}{4.9}} + \frac{h}{343} = 3$

$h = 40.7$ m