A stone is dropped into a well while splash is heard after 4.5 seconds. Another stone is dropped with an initial velocity, v and the splash is heard after 4 seconds. If the velocity of the
sound is 336m/s, determine the initial velocity of second stone
Answers
The time taken to hear the splash is 4 seconds from the moment the stone was released inside the well. If this is what is considered, then you have to account the velocity of the stone and some other related variables and solve for them using the SUVAT equations. But since this data is not given, I’m assuming the second interpretation is what is intended (which is given below)
The time taken to hear the splash is 4 seconds from the moment the stone hits the water and gives rise to the splash. This means that the splash sound takes 4 seconds to reach the listener. For all intents and purposes, I’ll be solving the question assuming this is the interpretation.
Now, the only formula required is speed = distance/time. Since time is four seconds and speed of sound is given to be 350 m/s, distance here represents the depth of the well, because the sound must have travelled this entire distance from the surface of the water to reach the listener, which is also the depth of the well.
distance=350∗4=1400m.
So the depth of the well is 1,400 meters.
Note that I have made some assumptions above to obtain this answer using the second interpretation:
It is assumed that the point from which the stone is released and the listener is located, is the exact point that indicates the opening of the well. If this is not the case, then we have to account for the extra distance between the opening of the well and the listener/point of release and subtract it from the distance that we have obtained (assuming that the distance between the opening and the listener/point of release is given).
It is assumed that the surface of the water is the bottom of the well. If this is not the case (which in reality isn’t), then the actual depth of the well is calculated by taking the distance obtained using the time and speed of sound (which I have got above), and adding it to the depth of water in the well, assuming the depth of water is given.
Note: for those interested, the way to solve this using the first interpretation is given below:
u = 0 (initial velocity is zero since the stone is assumed to start from rest)
a = 9.8 m/s^2 (acceleration due to gravity in the positive direction)
Therefore, using the equation S = ut + 1/2at^2,
2S/a = t^2 or sqrt(2S/9.8) = t (Time taken for the stone to cover the distance)
Now, 4 = sqrt(2S/9.8) + S/350 (here S/350 is the time taken for the splash sound to return back)