**Introduction to Falling Bodies**

It is known that, near the surface of the earth, a body falls with acceleration (due to gravity) of about 32 ft/sec ^{2} . If we let *h* ( *t* ) be the height of the object at time *t* (measured in seconds), then our information is that

*h″* ( *t* ) = −32.

Observe the minus sign to indicate that height is decreasing.

Now we will do some organized guessing. What could *h′* be? It is some function whose derivative is the constant −32. Our experience indicates that polynomials decrease in degree when we differentiate them. That is, the degree goes from 5 to 4, or from 3 to 2. Since, *h″* is a polynomial of degree 0, we therefore determine that *h′* will be a polynomial of degree 1. A moment’s thought then suggests that *h′* ( *t* ) = −32 *t* . This works! If *h′* ( *t* ) = −32 *t* then *h″* ( *t* ) = [ *h′* ( *t* )′ = −32. In fact we can do a bit better. Since constants differentiate to zero, our best guess of what the velocity should be is *h′* ( *t* ) = −32 *t* + *υ* _{0} , where *υ* _{0} is an undetermined constant.

Now let us guess what form *h* ( *t* ) should have. We can learn from our experience in the last paragraph. The “antiderivative” of −32 *t* (a polynomial of degree 1) should be a polynomial of degree 2. After a little fiddling, we guess −16 *t* ^{2} . And this works. The antiderivative of *υ* _{0} (a polynomial of degree 0) should be a polynomial of degree 1. After a little fiddling, we guess *υ* _{0} *t* . And this works. Taking all this information together, we find that the “antiderivative” of *h′* ( *t* ) = −32 *t* + *υ* _{0} is

Notice that we have once again thrown in an additive constant *h* _{0} . This does no harm:

*h′* ( *t* ) = [−16 *t* ^{2} ]′ + [ *υ* _{o} *t* ]′ + [ *h* _{0} ]′ = −32 *t* + *υ* _{0} ,

just as we wish. And, to repeat what we have already confirmed,

*h″* ( *t* ) = [ *h′* ( *t* )]′ = [−32 *t* ]′ + [ *υ* _{o} ]′ = −32.

We now have a general formula (namely (†)) for the position of a falling body at time *t* . [Recall that we were first introduced to this formula in Section 2.6.] See Fig. 3.17.

**Fig. 3.17**

Before doing some examples, we observe that a falling body will have initial velocity 0. Thus

0 = *h′* (0) = −32 · 0 + *υ* _{0} .

Hence, for a falling body, *υ* _{0} = 0. In some problems we may give the body an initial push, and then *υ* _{0} will not be zero.**
**

**Example 1**

Suppose that a falling body hits the ground with velocity −100 ft/sec. What was the initial height of the body?

**Solution 1**

With notation as developed above, we know that velocity is given by

*h′* ( *t* ) = −32 *t* + 0.

We have taken *υ* _{0} to be 0 because the body is a falling body; it had no initial push. If *T* is the time at which the body hits the ground, then we know that

−100 = *h′* ( *T* ) = −32 · *T* .

As a result, *T* = 25/8 sec.

When the body hits the ground, its height is 0. Thus we know that

0 = *h* ( *T* ) = *h* (25/8) = −16 · (25/8) ^{2} + *h* _{0} .

We may solve for *h* _{0} to obtain

Thus all the information about our falling body is given by

At time *t* = 0 we have

Thus the initial height of the falling body is 625/4 ft = 156.25 ft.

Notice that, in the process of solving the last example, and in the discussion preceding it, we learned that *h* _{0} represents the initial height of the falling body and *υ* _{0} represents the initial velocity of the falling body. This information will be useful in the other examples that we examine.

**Example 2**

A body is thrown straight down with an initial velocity of 10 feet per second. It strikes the ground in 12 seconds. What was the initial height?

**Solution 2 **

We know that *υ* _{0} = −10 and that *h* (12) = 0. This is the information that we must exploit in solving the problem. Now

*h* ( *t* ) = −16 *t* ^{2} − 10 *t* + *h* _{0} .

Thus

0 = *h* (12) = −16 · 12 ^{2} − 10 · 12 + *h* _{0} .

We may solve for *h* _{0} to obtain

*h* _{0} = 2424 ft.

The initial height is 2424 feet.

**You Try It:** A body is thrown straight up with initial velocity 5 feet per second from a height of 40 feet. After how many seconds will it hit the ground? What will be its maximum height?

**Example 3**

A body is launched straight up from height 100 feet with some initial velocity. It hits the ground after 10 seconds. What was that initial velocity?

**Solution 3**

We are given that *h* _{0} = 100. Thus

*h* ( *t* ) = −16 *t* ^{2} + *υ* _{0} *t* + 100.

Our job is to find *υ* _{0} . We also know that

0 = *h* (10) = −16 · 10 ^{2} + *υ* _{0} · 10 + 100.

We solve this equation to find that *υ* _{0} = 150 ft/sec.

**You Try It:** On a certain planet, bodies fall with an acceleration due to gravity of 10 ft/sec ^{2} . A certain body is thrown down with an initial velocity of 5 feet per second, and hits the surface 12 seconds later. From what height was it launched?

Find practice problems and solutions for these concepts at: Applications of the Derivative Practice Test.