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November 26, 2015
Posted by **Jaleel** on Tuesday, October 30, 2007 at 6:41pm.

- Math -
**tchrwill**, Tuesday, October 30, 2007 at 7:04pmPlease explain how to figure out the following 7th grade math problem: If a rubber ball bounces exactly 1/2 the height from which it is dropped; and it is dropped from the top of a building that is 64 meters tall, how high will the ball bounce on its eighth bounce?

64(1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + etc.

Get the idea?

- Math -
**Doug**, Wednesday, October 31, 2007 at 3:24pmIn algebra terms, you could look at the problem like this:

A. Set your variables. Do this by thinking about which parts of the problem can change. For this problem, there is the height of the final bounce, the height of the building, and the number of bounces.

1) h = Height of final bounce

2) b = Height of building

3) n = Number of bounces

B. Look for the pattern (see tchrwill's example above).

Looks like in the fraction, the numerator "1" stays the same. The denominator changes though, as follows.

1 bounce = 1/2

2 bounces = 1/4

3 bounces = 1/8

4 bounces = 1/16

etc.

If you look closely, denominator seems to be powers of 2, such as 2exp(1) = 2; 2exp(2) = 4; 2exp(3) = 8; 2exp(4) = 16; and so on.

C. Try making a formula.

Based on the numbers above, the formula should probably be:

h = b x [1/2exp(n)]

Just think through this formula step by step and you'll see how I got that.

D. Test it (see if the formula works).

For the eighth bounce, we have

h = b x [1/2exp(n)]

h = 64m x [1/2exp(8)]

Note that 2exp(8) = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2. Multiply those out and you get 256.

So,

h = 64m x [1/256]

Note that 1/256 approximately equals 0.004. Get this by long division or by calculator.

Finally,

h = 64m x 0.004

h = 0.256m

So on the eighth bounce, the ball should bounce about 256mm.

You should also plug in n = 1 and n = 2 just to satisfy yourself that the final answer is right.

Hope that helps!

Doug Jobes

- Math -
**gman**, Wednesday, May 22, 2013 at 4:27amyour formula is not right doug because there is already a formula which says b(common ratio exp(n-1)=h of n..