Falling Bodies Help

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

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.