rubber ball dropped

**gumi** · May 31st 2008, 05:08 PM

A rubber ball dropped from a height of 10 meters, rebounded 3/4 of it's height from which it fell. If it continued to bounce and rebound so that each new height was 3/4 the previous height, how far did the ball travel before hitting the ground 20 times?

**galactus** · May 31st 2008, 05:34 PM

The nth partial sum of a geometric series is

$S_{n}=a_{1}\frac{1-r^{n}}{1-r}$

Where $a_{1}=\frac{15}{2}, \;\ r=\frac{3}{4}$

The infinite series can be gotten from

$10+2\left[7.5+7.5(3/4)+7.5(3/4)^{2}+7.5(3/4)^{3}+.....\right]$

$=10+2\left(\frac{\frac{15}{2}}{1-\frac{3}{4}}\right)$

**mathaddict** · November 5th 2008, 06:44 AM

y can't we take the sum of distance travelled by the ball after hitting the ground 20 times ? In what case , can we consider that it is an infinite series ?

**HallsofIvy** · November 6th 2008, 04:02 AM

No, this is not an infinite series. Nor is it a geometric series!

First the ball drops 10 m. Then it rebounds (3/4)(10) m. Then it drops (3/4)(10)m, then rebounds $(3/4)^2 10$ m., then drops $(3/4)^2 10$ m. Each distance, after the first drop is doubled- once up then down. The total distance is $10+ 2(10)(3/4)+ 2(10)(3/4)^2+ \cdot\cdot\cdot+ 2(10)(3/4)^n$ . Notice that it has bounced once after the first drop, twice at the second drop, ... and when the exponent is n, has bounced n+1 times. The total distance traveled at the 20th bounce is $D= 10+ 20(3/4)+ 20(3/4)^2+ \cdot\cdot\cdot+ 20(3/4)^19$

That is NOT a geometric series because the first (n=0) term is 10, not 20. Fortunately, we can fix that by adding 10 to both sides: $D+10= 20+ 20(3/4)+ 20(3/4)^2+ \cdot\cdot\cdot+ 20(3/4)^19$

Now use the formula $\sum_{n=0}^N= \frac{a(1- r^{N+1})}{1- r}$ .

$D-10= \frac{20(1- (3/4)^{20})}{1- 3/4}= 80(1- (3/4)^{20})= 79.75$ m, approximately so D= 69.75 m, approximately.

(The total distance after an "infinite" number of bounces would be 70 m.)

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