Question

A stone is dropped into and the sound of impact of the stone with water is heard after $2.053\;s$ of the release of stone from the top. If the acceleration due to gravity is $980\;cms^{-2}$ and the velocity of sound in air is $350ms^{-1}$, then the depth of the well is:
A. 7m
B. 19.40m
C. 700m
D. 1960m

Answer 1

Hint:A good point to start would be to obtain expressions for time taken by the stone to hit the water and the time taken by the sound of impact to reach the surface, as the time specified in the question is the sum of the two. Then once you perform the appropriate substitutions you should arrive at a quadratic equation for one of the times. Then solve the action and consider the factor that yields a positive value, and substitute this time back in one of your equations that relates this time to the depth of the well, or the height travelled by the sound wave, both are the same.

Formula used:
Equation of motion to determine the depth travelled by the stone:
$h = \dfrac{1}{2}gt^2$, where g is the acceleration due to gravity and t is the time taken to travel to that depth.

Complete answer:
Let us begin by deconstructing the information given to us for a better understanding.
Let the time taken by the stone to reach the water be $t_1$ and let the time taken by the sound of impact to come to the surface be $t_2$.
Let h be the depth of the wall travelled by the stone till it hits the water. From the kinematic equation of motion, under the acceleration due to gravity, we have
$h = \dfrac{1}{2}g t_1^2$ since the stone is just dropped into the well.
Given that the velocity of the sound is $350ms^{-1}$, in the time $t_2$ taken by the sound to come to the surface it travels the same height h.
$h = 350 \times t_2 \Rightarrow t_2 = \dfrac{h}{350}$ but we found h to be $h = \dfrac{1}{2}g t_1^2$ as well. Substituting this, we get:
$t_2 = \dfrac{\dfrac{1}{2}g t_1^2}{350} = \dfrac{gt_1^2}{700}$
From the question, we have $t_1 +t_2 = 2.056$ which now becomes $ t_1 + \dfrac{gt_1^2}{700} = 2.053$
Substituting $g = 980\;cms^{-2} = 9.8\;ms^{-2}$ we get
$t_1 + \dfrac{9.8t_1^2}{700} = 2.053$
$\Rightarrow 700t_1 + 9.8t_1^2 = 1437.1 \Rightarrow 9.8t_1^2 + 700t_1 -1437.1 = 0$
Since the decimal makes it difficult to factorize the equation, let us multiply the whole equation by 10 for easing our calculation
$\Rightarrow 98t_1^2 + 7000t_1 -14371 = 0$
$\Rightarrow 14t_1^2 + 1000t_1 -2053 = 0
$\Rightarrow (t_1-1.99) (t_1+73.4)=0 $
Since negative time does not hold any significance to us, we only consider the factor that yields a positive time:
$t_1 -1.99 = 0 \Rightarrow t_1 =1.99\;s$
Therefore, the depth of the well will be:
$ h = \dfrac{1}{2}g t_1^2 = \dfrac{1}{2} \times 9.8 \times (1.99)^2 = $\dfrac{9.8 \times 3.96}{{2}}$ = 19.40\;m$
Therefore, the correct option would be B. 19.40m

Note:
The key note in such problems is to ensure that you are able to grasp all the parameters that are given, direct or indirect. For instance, it was up to us to understand that if a sound was heard after $t\;s$ it meant that the sound wave had to travel up from the well to reach us and had to travel the same height as the depth at which the stone intercepted the water. It is thus very important to catch the things you need from a problem to go about finding the accurate solution.
Also note that since the stone was just dropped into the well it carried no launch or initial velocity. In case it was projected or thrown into the well, the equations would be entirely different and more data would be required to treat the situation as that of a projectile motion.
And remember that if the stone was thrown upwards, then the acceleration due to gravity would possess a negative sign instead.