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Simple throwing ball up problem

  1. Dec 15, 2007 #1
    Simple throwing ball "up" problem

    1. The problem statement, all variables and given/known data
    Ball is thrown vertically with upward velocity of 18m/s when it reaches 1/4 of its maximum height above its launch point. Find initial launch velocity of ball.


    2. Relevant equations
    0.5mv2, mgh, v=d/t


    3. The attempt at a solution

    The 1/4 of the maximum height is throwing me off.
    I tried using kinematics equations to first solve for the first part of the throw (the 1/4)
    and then use final velocity of zero at the max height using deceleration of gravity to find this height.
    I'm stumped, any help is appreciated.
    Thanks.
     
  2. jcsd
  3. Dec 15, 2007 #2

    Dale

    Staff: Mentor

    I would do this with conservation of energy. Set your h=0 point at the 1/4 point (where v=18m/s)
     
  4. Dec 15, 2007 #3

    robphy

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    Gold Member

    v is not generally equal to d/t.
    While you can use energy conservation, in case you haven't gotten that far... or wish to use something more kinematical, you can use a formula (resembling the energy conservation approach) derived from the constant-acceleration position-vs-time and velocity-vs-time equations.
     
  5. Dec 15, 2007 #4
    Ok thanks, I found the height to be 32.8 m (?) at the quarter point, then set KE + PE final to = KE initial and got 25 m/s for an initial velocity. The answer is 21 m/s though. Not sure where I went wrong.
     
  6. Dec 16, 2007 #5
    Your height value is incorrect. I'd use conservation of energy to find the height (you don't need to set the height at the 1/4 mark to zero, I find this adds an unnecessary step; just set it equal to 1/4h). What is your energy conservation expression? Once you have the correct value of the height, it's merely a kinematics problem.
     
  7. Dec 16, 2007 #6

    rl.bhat

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    If h is the maximum height, you can wright two equation.
    v1^2 - vi^2 = -2gh/4...(1)
    0 - vi^2 =-2gh ....(2) solve these equation and find vi
     
    Last edited: Dec 16, 2007
  8. Dec 16, 2007 #7
    rl.bhat, those equations are wrong... should be v1^2-vi^2 = -2gh/4....
     
  9. Dec 16, 2007 #8
    My conseravtion of energy expression is .5mv^2 + mgh = .5mv^2
    This leaves me with two variables, so I am not sure what is next.
    Thanks.
     
  10. Dec 16, 2007 #9

    Hootenanny

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    Gold Member

    Perhaps this would be helpful;
     
  11. Dec 16, 2007 #10
    Set it up so your initial velocity is 18m/s. What is the height when the velocity is 18 m/s? What is the velocity when the ball is at maximum height? Once you know the height, the rest will fall into place...
     
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