Solve the applied problem.Tom drops a rock into a well and 3 seconds later hears the sound...

Solve the applied problem.

Tom drops a rock into a well and 3 seconds later hears the sound of its splash. How deep is the well? [Make the same assumptions about falling rocks and sound as in Example 1.]

EXAMPLE 1

A stone is dropped from the top of a cliff. Five seconds later, the sound of the stone hitting the ground is heard. Use the fact that the speed of sound is 1100 feet per second and that the distance traveled by a falling object in t seconds is 16t2 feet to determine the height of the cliff.

SOLUTION If t is the time it takes the rock to fall to the ground, then the height of the cliff is 16t2 feet. The total time for the rock to fall and the sound of its hitting ground to return to the top of the cliff is 5 seconds. So the sound must take 5 − t seconds to return to the top, as indicated in Figure 1. Therefore

The quadratic formula and a calculator show that

Since the time must be a number between 0 and 5, we see that t ≈ 4.6812 seconds. Therefore the height of the cliff is

 16t2 = 16(4.6812)2 ≈ 350.6 feet.

Figure 1