Question

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.

Answer 1

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

Answer 2

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

Answer 3

Could somebody please explain how they got that?

Answer 4

Why n-1?

Answer 5

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"

Answer 6

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

Answer 7

@Zas does this help?

Answer 8

I think so, thank you!

Answer 9

You're welcome. :]