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.
Add your own comment
Today on Education.com
WORKBOOKSMay Workbooks are Here!
WE'VE GOT A GREAT ROUND-UP OF ACTIVITIES PERFECT FOR LONG WEEKENDS, STAYCATIONS, VACATIONS ... OR JUST SOME GOOD OLD-FASHIONED FUN!Get Outside! 10 Playful Activities
- Kindergarten Sight Words List
- The Five Warning Signs of Asperger's Syndrome
- What Makes a School Effective?
- Child Development Theories
- Why is Play Important? Social and Emotional Development, Physical Development, Creative Development
- 10 Fun Activities for Children with Autism
- Test Problems: Seven Reasons Why Standardized Tests Are Not Working
- Bullying in Schools
- A Teacher's Guide to Differentiating Instruction
- First Grade Sight Words List