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anonymous

  • 5 years ago

On the moon the acceleration due to gravity is 1.6 metres per second squared (approximately 1 6 th of the value on earth). Standing on top of a ladder, 5 metres up, the astronaut throws a ball up vertically into the air with velocity 2m/s. How long does it take to reach the ground? How long would it take to reach the ground if the same experiment were done on earth?

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  1. anonymous
    • 5 years ago
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    look in google

  2. anonymous
    • 5 years ago
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    lol

  3. anonymous
    • 5 years ago
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    THAT DOESN'T HELP!

  4. anonymous
    • 5 years ago
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    I have an answer, but I am not entirely sure if it's correct or not. I'll write it down anyway.

  5. anonymous
    • 5 years ago
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    go for it

  6. anonymous
    • 5 years ago
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    \[\Delta x=v_i t+{1 \over 2}at^2\], where delta x=-5 (the change in the position), v_i=2m/s and a=-1.6m/s^2. Substituting them in the equation gives: \[-5=2t+{1 \over 2}(-1.6)t^2 \implies 0.8t^2-2t-5=0\]

  7. anonymous
    • 5 years ago
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    Solve the quadratic equation for t, you get one positive value for t. That's t=4.05 sec

  8. anonymous
    • 5 years ago
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    Do you have the answer, or choices?

  9. anonymous
    • 5 years ago
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    i only got to having to use the quadratic equation, i didn't realise i had to use the quadratic equation

  10. anonymous
    • 5 years ago
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    You can just do it in your calculator.

  11. anonymous
    • 5 years ago
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    yeah i just had to recognise that cause i think i'm meant to show working

  12. anonymous
    • 5 years ago
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    the quadratic formula is your personal picture yo :P

  13. anonymous
    • 5 years ago
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    yeah, cause i m having maximum trouble with it :)

  14. anonymous
    • 5 years ago
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    LOL!

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spraguer (Moderator)
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