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stracyl

  • 2 years ago

an object is thrown directly up with a velocity of 20.0 m/s and a d0=0. Determine how long it takes to get to the maximum height of 24.0 m

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  1. iNeedTheHelp
    • 2 years ago
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    Well we know the maximum height occurs when the velocity is at 0 (initial velocity)(time) = (0.5)a(t^2) Diviide both sides by 't velocity = at/2 time = 2v/a = (2)(20.0) / (9.81 m/s/s) time = 4.077sec

  2. nubeer
    • 2 years ago
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    hmm i think something wrong in there.. @iNeedTheHelp can you tell which formula you have used?

  3. iNeedTheHelp
    • 2 years ago
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    let me see hmm

  4. nubeer
    • 2 years ago
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    if i am right u have used s=ut+0.5at^2.. right?

  5. iNeedTheHelp
    • 2 years ago
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    Yeah is there a flaw in the equation?

  6. nubeer
    • 2 years ago
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    well if you have use this formula , in it u=initial velocity. and in question initial velocity is given.

  7. iNeedTheHelp
    • 2 years ago
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    Yes I already know that but what is wrong with this equation and sorry I'm kind of tired

  8. nubeer
    • 2 years ago
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    its ok.. well your equation will be then s = 24t+0.5at^2

  9. nubeer
    • 2 years ago
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    you missed 24t in your equation.

  10. iNeedTheHelp
    • 2 years ago
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    ohhh wow thats a big one too sorry but @nubeer you can help him because I am going to bed

  11. nubeer
    • 2 years ago
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    sure no problem :) have sweet dreams ok @stracyl formula is this s=ut+0.5at^2.. s=24, u = 20 a=9.81 24 = 20t+0.5(9.81)t^2 0.5(9.81)t^2 +20t -24 = 0 use quadratic formula and find the roots of t

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