Question

A pebble is dropped into a deep well and 3s later the sound of a splash is heard as the pebble reaches the bottom of the well. The speed of sound in air is about 340 m/s. How long will it take fir the pebble to hit the water? How long does is take for the sound to reach the observer? What is the depth of the well? (I need the equation)

Answer 1

Answer:

$2.89\ \text{s}$

0.11 seconds

$37.4\ \text{m}$

Step-by-step explanation:

$t_1$ = Time taken by the pebble to hit the water

$t_2$ = Time taken by the sound to travel to the observer

$t_1+t_2$ = Time taken by the observer to listen to the splash after stone is thrown = 3 s

$t_1=3-t_2$

a = g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

$u_s$ = Speed of sound = 340 m/s

The distance traveled by the pebble

$s=ut+\dfrac{1}{2}at_1^2\\\Rightarrow s=0+\dfrac{1}{2}\times9.81(3-t_2)^2\\\Rightarrow s=4.905(9+t_2^2-6t_2^2)$

Distance traveled by the sound

$s=u_st\\\Rightarrow s=340\times t_2$

The distance traveled by the sound and the pebble is equal

$4.905(9+t_2^2-6t_2)=340\times t_2\\\Rightarrow 9+t_2^2-6t_2=\dfrac{340}{4.905}t_2\\\Rightarrow 9+t_2^2-6t_2=69.32t_2\\\Rightarrow t_2^2-75.32t_2+9=0\\\Rightarrow t_2=\frac{-\left(-75.32\right)\pm \sqrt{\left(-75.32\right)^2-4\times \:1\times \:9}}{2\times \:1}\\\Rightarrow t_2=75.2,0.11$

Since $t_1+t_2=3$ the value of $t_2$ cannot be greater than 3.

So, time taken by the sound to reach the observer is 0.11 seconds.

Time taken by the pebble to hit the water is $t_1=3-t_2=3-0.11=2.89\ \text{s}$

Depth of the well is $s=340t_2=340\times 0.11=37.4\ \text{m}$