## anonymous one year ago A ball is thrown vertically upwards from the ground. It rises to a height of 10 m and then falls and bounces. After each bounce it rises vertically to 2/3 of the height from which it fell Find the height to which the ball bounces after the nth impact with the ground. I will put the answer I got now.

1. anonymous

$10 \times (\frac{ 2 }{ 3 })^{k-1}$ Thats what I got

2. anonymous

The answer is $10 \times (\frac{ 2 }{ 3 })^{n}$

3. anonymous

Could somebody please explain how they got that?

4. kittiwitti1

Why n-1?

5. baru

if the answer was (k-1), then after the first bounce the ball would rise by 10* (2/3)^0 which is equal to 10. thats why its "n"

6. kittiwitti1

$n\text{ is the number of times it bounces off the ground, so that becomes}$$\left(\frac{2}{3}\right)^{n}$$\text{multiply that by the original height bounced of 10 meters and}$$10\times\left(\frac{2}{3}\right)^{n}$Basically it is saying "imagine 10 m per bounce scenario and then multiply 10 m by the amount of bounces, then 2/3 that"

7. kittiwitti1

@Zas does this help?

8. anonymous

I think so, thank you!

9. kittiwitti1

You're welcome. :]