1951 AMC 12 Problem 39

Question

A stone is dropped into a well and the report of the stone striking the bottom is heard

7.7

seconds after it is dropped. Assume that the stone falls

16t^2

feet in

t

seconds and that the velocity of sound is

1120

feet per second. The depth of the well is:

784

ft.

342

ft.

1568

ft.

156.8

ft.

None of these

The total time is the falling time plus the sound’s return time

If the falling time is

t

, solve

t+\dfrac{16t^2}{1120}=7.7

If the stone falls for

t

seconds, the depth is

16t^2,

and the sound takes

\frac{16t^2}{1120}=\frac{t^2}{70}

seconds to return. Hence

t+\frac{t^2}{70}=7.7,

or

t^2+70t-539=0.

The positive root is

t=7,

giving depth

16(7^2)=784

feet.

Thus, the correct answer is A.

Problem 39 in other years