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.

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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.

Mathematics
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\[10 \times (\frac{ 2 }{ 3 })^{k-1}\] Thats what I got
The answer is \[10 \times (\frac{ 2 }{ 3 })^{n}\]
Could somebody please explain how they got that?

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Other answers:

Why n-1?
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"
\[n\text{ is the number of times it bounces off the ground, so that becomes}\]\[\left(\frac{2}{3}\right)^{n}\]\[\text{multiply that by the original height bounced of 10 meters and}\]\[10\times\left(\frac{2}{3}\right)^{n}\]Basically it is saying "imagine 10 m per bounce scenario and then multiply 10 m by the amount of bounces, then 2/3 that"
@Zas does this help?
I think so, thank you!
You're welcome. :]

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