Question

A stone is dropped into a well and the sound of impact of stone on the water is heard after 2.056sec of the release of stone from the top. If acceleration due to gravity is $980cm/{{\sec }^{2}}$ and velocity of sound in air is $350m/s$ , calculate the depth of the well.

Answer 1

The total time taken for the sound of impact to be heard from the moment of its release is given. You could make use of Newton's equation of motion to express depth of the well H in terms of the time taken for stone to reach the bottom. Now find the time taken for the sound of impact to reach the top with given velocity of sound in air and the substitute the above expression for H. From the given total time you could get a quadratic equation in terms of ${{t}_{1}}$ , solving that and substituting accordingly, you will get the depth of the well.

Formula used:
Newton’s equation of motion,
$s=ut+\dfrac{1}{2}g{{t}^{2}}$

Complete step-by-step answer:
In the question, we are given a stone that is dropped into the well. The sound of impact of the stone when it hits the water is heard after 2.056sec of its release from top. We are given the acceleration due to gravity and velocity of sound in air as $980cm/{{s}^{2}}$ and $350m/s$ respectively. We are asked to find the depth of the well using the given information.

Let the time taken by the stone to drop to the bottom of the well be ${{t}_{1}}$ and the height of the well be H, then, the acceleration on the stone that is dropped will be that due to gravity.
$a=g=9.8m/s$
From Newton’s Laws of motion,
$s=ut+\dfrac{1}{2}g{{t}^{2}}$
For a body dropped, the initial velocity u = 0,
$H=\dfrac{1}{2}g{{t}_{1}}^{2}$ ……………………………………….. (1)
We know that,
$\dfrac{s}{t}=v$
We are given the speed of sound in air, so, let the time taken for the sound of impact to travel up the height of the well be ${{t}_{2}}$ , then,
$\dfrac{H}{{{t}_{2}}}=350m/s$
Substituting for H from (1),
${{t}_{2}}=\dfrac{\dfrac{1}{2}g{{t}_{1}}^{2}}{350}$
${{t}_{2}}=\dfrac{g{{t}_{1}}^{2}}{700}$ ………………………………………………. (2)
We are also given the total time taken for the sound of impact to reach the top from its release, so,
$T={{t}_{1}}+{{t}_{2}}=2.056s$
Substituting (2),
${{t}_{1}}+\dfrac{g{{t}_{1}}^{2}}{700}=2.056$
$\Rightarrow 7000{{t}_{1}}+98{{t}_{1}}^{2}=14392$
$\Rightarrow 14{{t}_{1}}^{2}+1000{{t}_{1}}-2056=0$
$\Rightarrow 14{{t}_{1}}\left( {{t}_{1}}-2 \right)+1028\left( {{t}_{1}}-2 \right)=0$
$\Rightarrow \left( {{t}_{1}}-2 \right)\left( 14{{t}_{1}}+1028 \right)=0$
So, ${{t}_{1}}$ can be either
${{t}_{1}}=2$
Or,
${{t}_{1}}=-\dfrac{1028}{14}$
Since ${{t}_{1}}$ here is the time taken, it can’t be negative. So,
${{t}_{1}}=2s$ ……………………………………………….. (3)
Substituting (3) in (1),
$H=\dfrac{1}{2}\left( 9.8 \right){{\left( 2 \right)}^{2}}$
$\therefore H=19.6m$
Hence, we got the depth of the well as,
$H=19.6m$

Note: Solving this kind of problem is pretty simple if you really understand the situation given. Here, all you have to understand is that the total time taken for the sound of impact to reach the top will be the sum of the time taken for the stone to reach the bottom and the time taken by the sound to reach the top. Knowing this and doing the simple necessary substitutions you will get the answer. So, always try to imagine what is really happening while reading the question.