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Max height-2 balls-1 straight up-the other horizontal at an angle? yikes!

  1. Sep 13, 2008 #1
    1. The problem statement, all variables and given/known data
    A ball is thrown straight upward and returns
    to the thrower’s hand after 4 s in the air. A
    second ball is thrown at an angle of 67◦ with
    the horizontal.
    The acceleration of gravity is 9.8 m/s2 .
    At what speed must the second ball be
    thrown so that it reaches the same height as
    the one thrown vertically? Answer in units of
    m/s.


    3. The attempt at a solution

    *****I've been working on this problem for a few days. Here's what I have so far:

    This is a 2-d question. First I need to find the maximum height of the first ball (thrown straight up) then I can plug that into the second ball's equations and find the Vo(x) of the second ball.
    **Will the ball that's being thrown straight up have it's maximum height at 2s (half way between the total time?)
    So I would have Y=0 + 0(1)+ (+)4.9(1^2).
    Then Height would be +4.9? I think that gravity COULD be + because the ball is going up...but I don't know if that even makes sense.

    Then I would put the Y into my second ball's equations:
    1.) Y EQN: 4.9=0+Vo(y)sin(67)(t)+4.9(t^2)
    2.) X EQN: x=0 + Vo(x)cos(67)(t)+ 0
    3.) Vy EQN: 0=Vo(y) + -9.8(t)
    4.) Vx EQN: Vx=Vo(x) + 0

    **Substitute and solve for t: t=.___s
    **Plug t into X EQN: x=Vo(x)cos(37)(.__s)
    Now I am missing x and the magnitude.

    I really don't know what to do with this problem. Any help is greatly appreciated!! Thank you! :)
     
  2. jcsd
  3. Sep 13, 2008 #2

    Kurdt

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    Staff Emeritus
    Science Advisor
    Gold Member

    You need to find the initial vertical velocity the first ball has. That will be the same vertical velocity your second ball has.
     
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