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anonymous

  • 5 years ago

If a 65kg woman is driving off a 10 meter platform, what is her speed as she hits the water?

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  1. anonymous
    • 5 years ago
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    more information is required to answer this question like her initial speed(or velocity),acceleration,frictional force(if any) etc

  2. anonymous
    • 5 years ago
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    No it isn't. Assuming no air resistance, you just use: \[v^2=u^2+2as\]Her initial vertical velocity is zero so this term disappears, you know s=10m and a=9.81ms^-2

  3. anonymous
    • 5 years ago
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    the woman is driving off the platform..how can her inital velocity be zero?

  4. anonymous
    • 5 years ago
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    I agree with bendt ans its correct

  5. anonymous
    • 5 years ago
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    Her initial vertical velocity is zero, not her horizontal velocity.

  6. anonymous
    • 5 years ago
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    before driving off the plateform her speed is zeroo

  7. anonymous
    • 5 years ago
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    At the instant she leaves the platform, the vertical component of her velocity vector is zero.

  8. anonymous
    • 5 years ago
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    mmm...but isnt the final speed affected by both vertical and horizontal components?

  9. anonymous
    • 5 years ago
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    horizontal component doesnot play any role in vertical velocity

  10. anonymous
    • 5 years ago
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    Yes, Vids93, but that's not what the question's asking. Think of her walking off the platform.

  11. anonymous
    • 5 years ago
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    i see the qs says what is the 'speed' as the woman hits the water!

  12. anonymous
    • 5 years ago
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    Yes but it clearly doesn't want you to take into account the horizontal velocity because, as you said, you would need more information.

  13. anonymous
    • 5 years ago
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    Thanks bendt..:)

  14. anonymous
    • 5 years ago
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    No problem.

  15. anonymous
    • 5 years ago
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    What would the u^2 stand for?

  16. anonymous
    • 5 years ago
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    Initial velocity.

  17. anonymous
    • 5 years ago
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    Ok thank you!

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