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Homework Help: What speed must the ball be thrown

  1. Aug 24, 2011 #1
    1. The problem statement, all variables and given/known data

    What speed must a ball be thrown vertically to reach a max height of 70m?

    2. Relevant equations

    3. The attempt at a solution

    I originally tried to find the velocity by the fact that the velocity of the ball must be zero at a height of 70 m, accounting for 9.8 m/s^2 acceleration due to gravity (which is negative in this case)

    So, if the velocity is decreased by 9.8 m/s every second, and reaches 70 m at zero, then I could find that if I knew the time that the ball is in the air.

    To find the time, I could use x=xo + vot + (1/2)at^2, but then I need initial velocity!

    What formula am I missing?
  2. jcsd
  3. Aug 24, 2011 #2
    Try using the first equation of motion.
    You know that at the maximum height the velocity is zero, using that you can find the value of v0 (in terms of g and t).
    Then substitute it in second equation of motion. ;)
  4. Aug 24, 2011 #3
    I'm not sure I understand how I could apply that one. I have v, as 0, but the only other value I know is a at 9.8.

    0 = Vo + 9.8t

    What can that tell me?

  5. Aug 24, 2011 #4
    Oops!! You did a sign mistake :smile:
    What you should get is this:-

    For now, don't substitute g, when the equations are solved then you can substitute g.
  6. Aug 24, 2011 #5
    So Vo=gt. So replace all instances of Vo with gt in the time equation, right?

  7. Aug 24, 2011 #6
    Yes! :)
  8. Aug 24, 2011 #7
    Do you agree with t = 2.18?
  9. Aug 24, 2011 #8
    My answer comes out to be 3.74s when g=10 m/s^2.
    It dont come 2.18s too when i take g=9.8 m/s^2
  10. Aug 24, 2011 #9

    Doc Al

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    Staff: Mentor

    Since Vo is what you want to solve for, not really the time, wouldn't it make more sense to eliminate t instead?
  11. Aug 24, 2011 #10
    Going that way I get vo= 21.4. Sorry I am showing no work, on a cellphone.
  12. Aug 24, 2011 #11

    Doc Al

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    Staff: Mentor

    No, but I have no idea what you did wrong.
  13. Aug 24, 2011 #12
    Alright, guess ill just have to wait until I get home, sit down, and think about it.
  14. Aug 24, 2011 #13
    Here's my work for replacing Vo

    Given that Xo = 0, X = 70, Vo = gt

    [itex]x = x_{0} + v_{0}t + \frac{1}{2}gt^{2}[/itex]

    Replace instances of Vo with gt, Xo is 0. x is 70.

    [itex]70 = (gt)t + \frac{1}{2}gt^{2}[/itex]

    [itex]70 = gt^{2} + \frac{1}{2}gt^{2}[/itex]

    [itex]70 = 9.8t^{2} + 4.9t^{2}[/itex]

    [itex]70 = 17.4t^{2}[/itex]

    [itex]4.76 = t^{2}[/itex]

    t = 2.18
  15. Aug 24, 2011 #14

    Doc Al

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    Staff: Mentor

    Your last term has the wrong sign. The acceleration is -g, not g.
  16. Aug 24, 2011 #15
    Of course! Ball is thrown against gravity. Accounting for that, 3.77964473..

    Thanks a lot guys.
  17. Aug 24, 2011 #16
    37.42 for initial velocity?
  18. Aug 25, 2011 #17
    Yep you got it. :smile:
  19. Aug 25, 2011 #18

    Doc Al

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    Good. Now try doing as I suggest in post #9 and solve it without the unnecessary step of calculating the time.
  20. Aug 25, 2011 #19
    It wasn't unnessesary because part b asked for it anyway, but will do.
  21. Aug 25, 2011 #20

    Doc Al

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    Staff: Mentor

    It's worth doing. And in the process you'll derive a very useful kinematic formula that will enable you to solve for the velocity in one step. (Well worth committing to memory.)
  22. Aug 25, 2011 #21


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    Homework Helper

    Now that you have the answer, I can say:

    One thing to consider is that if the ball was instead dropped from 70m, it would reach the ground at the speed you sought [just down instead of up, that's why I said speed not velocity], and the time to reach the ground would have been part 2 of the question you mentioned.
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