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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.

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  1. anonymous
    • one year ago
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    \[10 \times (\frac{ 2 }{ 3 })^{k-1}\] Thats what I got

  2. anonymous
    • one year ago
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    The answer is \[10 \times (\frac{ 2 }{ 3 })^{n}\]

  3. anonymous
    • one year ago
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    Could somebody please explain how they got that?

  4. kittiwitti1
    • one year ago
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    Why n-1?

  5. baru
    • one year ago
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    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
    • one year ago
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    \[n\text{ is the number of times it bounces off the ground, so that becomes}\]\[\left(\frac{2}{3}\right)^{n}\]\[\text{multiply that by the original height bounced of 10 meters and}\]\[10\times\left(\frac{2}{3}\right)^{n}\]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
    • one year ago
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    @Zas does this help?

  8. anonymous
    • one year ago
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    I think so, thank you!

  9. kittiwitti1
    • one year ago
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    You're welcome. :]

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