Question

A man drops a rock into a well.

(a) The man hears the sound of the splash 2.20 s after he releases the rock from rest. The speed of sound in air (at the ambient temperature) is 336 m/s. How far below the top of the well is the surface of the water? (answer in meters)

(b) If the travel time for the sound is ignored, what percentage error is introduced when the depth of the well is calculated? (percentage)

Answer 1

There are two times to work out - the time for the rock to fall and the time for the sound to come back to the surface. The common factor in both is the distance from the top of the well to the water.

First consider the time taken for the rock to fall to the surface of the water, t1. Since u = 0 and a = g = 9.8m/s/s, the distance is given by

s = ut+0.5at^2 = 0 + 0.5*9.8*t1^2 = 4.9*t1^2

Secondly, consider the time for the sound to come back up, t2. Since v = s/t2 = 336, then
s = v*t2 = 336*t2

The distance is common, s = s, so putting these together,
4.9*t1^2 = 336*t2

Now we know that t1 + t2 = 2.2s, so
t2 = 2.2 - t1
Substituting this into the first equation,
4.9*t1^2 = 336*(2.2 - t1), or

4.9*t1^2 + 336*t1 - 739.2 = 0

using the quadratic formula,

t1 = (-336 +/- sqrt(336^2 + 4*4.9*739.2))/(2*4.9) = 2.13s (or -70.7 which is clearly not the answer)

Now we can find the distance from

s = 0.5*9.8*t1^2 = 4.9*2.13^2 = 22.23m (answer 1)

If t2 is ignored, then we would take t1 to be equal to 2.20s, so
s = 4.9*2.2^2 = 23.716m

The error is 23.716 - 22.23 = 1.486which, as a percentage of the correct answer, is

1.486*100/22.23 = 6.68% (answer 2)