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The speed which a ball must be thrown to reach 8m.

  1. Nov 9, 2015 #1
    1. The problem statement, all variables and given/known data

    An object of mass 0.2 kg is thrown vertically upwards from ground level and reaches a height of 8.0 m. Calculate: the speed with which it must be thrown.

    2. Relevant equations

    Vf2 – Vo2 = 2gH

    3. The attempt at a solution

    Vf2 – Vo2 = 2gH
    Vf = Final Velocity = 0 (Ball at maximum height)
    Vo = Initial Velocity.
    g = 9.8
    H =8m
    0 - Vo2 = 2 x -9.8 x 8 = -156.8

    This is my attempt so far but I'm getting a negative answer, can anyone show me where I'm going wrong?
     
  2. jcsd
  3. Nov 9, 2015 #2

    Hesch

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    Suggesting there is no air resistance ( no loss of energy ):

    ½*m*V02 = m*g*H →

    V02 = 2*g*H

    or you could say:

    V02 – Vf2 = 2*g*H ( Vf = 0 ).
     
  4. Nov 9, 2015 #3
    This is what I'm already doing but getting a negative number? Should I just be times by 9.8 not -9.8?
     
  5. Nov 9, 2015 #4

    Hesch

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    No, I have switched the order of velocities, V0 and Vf .
    This will change the sign of the result.
     
  6. Nov 9, 2015 #5

    Hesch

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    And you should use g = +9.8 m/s2.
    Otherwise the potential energy will be negative.
     
  7. Nov 9, 2015 #6
    so V02 – Vf2 = 2*g*H ( Vf = 0 ).
    0 - Vf2 = 2 x 9.8 x 8 = 156.8.

    is this correct?
     
  8. Nov 9, 2015 #7

    Hesch

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    No, your first line is correct, but the second is not.

    V02 is not equal to 0.
     
  9. Nov 9, 2015 #8
    V02 = 2*g*H

    so V02= 2 x 9.8 x 8 = 156.8
     
  10. Nov 9, 2015 #9

    Hesch

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    Yes.

    Remember units!
     
  11. Nov 9, 2015 #10
    156.8 m/s correct?
    and thank you
     
  12. Nov 9, 2015 #11

    Hesch

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    No, try to insert units in this calculation:

    2 x 9.8 x 8 = 156.8

    Will the unit of the result be m/s? Or will it be m2/s2?
    Why?
     
  13. Nov 9, 2015 #12

    Hesch

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    V02 = 2 × 9.8 [m/s2] × 8 [m] = 156.8 [m2/s2] →
    V0 = (156.8 [m2/s2]) = 12.52 [m/s]

    Always check the units. Then you can see if something is wrong in your calculation.
    Here the square root was missing.
     
  14. Nov 9, 2015 #13
    Thank you for all your help :) it is greatly appreciated
     
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