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