Question

a ball dropped from a height of 12 feet and returns to a height that is one-half of the height from which it fell. the ball continues to bounce half the height of the previous bounce each time. how far will the ball have traveled when it hits the ground for the fifth time?

Answer 1

Answer:

Step-by-step explanation:

If we set up a table with the x and y values with x being the number of drops and y being the height at that drop, it would look like this:

x |  0   1   2   3     4     5

y | 12   6  3  1.5  .75  .375

That gives you your answer, sure, but what if you had to find the height of the ball at bouce number 7?  We can find the model for this particular situation by using the coordinates.  We know it's not linear because the slopes are not the same between the coordinate points.  But if we graph the points from above it is clear that is exponential decay, of the form:

$y=a(b)^x$

We can solve for a and b using the first 2 coordinates.  Then we will test our equation by plugging in x = 2 and seeing if what we get back is a y value of 3.

Fitting in (0, 12) gives us:

$12=a(b)^0$

Anything raised to the power of 0 is equal to 1, so we just found out that

a = 12.

Now we can use that value of a along with the next coordinate pair to solve for b.

l$6=12(b)^1$ which is

6 = 12b so

$b=\frac{1}{2}$

Fitting those into the standard form equation gives us a model of:

$y=12(\frac{1}{2})^x$

Let's fill in an x value of 2.  If this is the correct model, we should get a y value back of 3:

$y=12(\frac{1}{2})^2$ simplifies to

$y=12(\frac{1}{4})$ so

y = 3

There you go!  But the answer for you is that at the 5th bounce, the ball is .375 feet high or 3/8 feet