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Finding initial velocity

  1. Sep 11, 2007 #1
    1. The problem statement, all variables and given/known data

    A baseball thrown at an angle of 65.0 degrees above the horizontal strikes a building 16.0 m away at a point 5.00 m above the point from which it is thrown. Ignore air resistance.

    2. Relevant equations
    Not sure if this equation works for the problem

    3. The attempt at a solution
    I understand projectile motion but I'm confused because the ball hits the wall at a point higher than the initial point from which it's thrown. From my calculation I got 14.4 m/s squared. Range times gravity divided by sin130. I'm not sure of any other way of solving the problem but I know the above is wrong. I can't seem to get any of the kinematic equations down to one variable either.
  2. jcsd
  3. Sep 11, 2007 #2


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    Science Advisor

    Well, start form the basics: there is no acceleration in the horizontal direction, an acceleration of -g= -9.8 m/s2 in the vertical direction.

    accy= -9.81 so vy= -9.8t+ v0sin(65) and then, taking the initial height to be y= 0, y= -4.9t2+ v0 sin(65) t= 5.

    accx= 0 so vx= v0cos(65) and then, taking the initial position to be x= 0, x= v0cos(65)t= 16.

    That gives you two equations to solve for the two parameters t and v0.
  4. Sep 11, 2007 #3

    Doc Al

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    Staff: Mentor

    That "range formula" is too specialized to use here. Instead, go back to basics: Write equations for vertical (y) and horizontal (x) positions as a function of time. You'll get two equations and two unknowns, which you can solve for V0 and time.

    (Yeah, what he said. :wink:)
  5. Sep 11, 2007 #4
    Ok so I solve for the y equation using a quadratic then I plug t into x? Doc Al when do you know when to use the range equation and when its not appropriate. I have a hard time logically figuring out which equation to use.
  6. Sep 11, 2007 #5

    Doc Al

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    Staff: Mentor

    "Range" means horizontal distance, so the "range equation" is only useful when the initial and final points are at the same height. Read this: Range of Trajectory

    Unless you have a pile of range problems to solve, it's not worth memorizing. More important is to know the basic kinematics of projectile motion and how to derive the range when you need it.
  7. Sep 11, 2007 #6
    arghhh, my algebras rusty how do i solve these variables, don't tell me the answer tho but lead me in the right direction
  8. Sep 11, 2007 #7


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    Homework Helper

    Solve for time in the horizontal displacement equation in terms of V0... then substitute that value for time, into the vertical displacement equation.
  9. Sep 11, 2007 #8
    t = 16/VOcos65 is that in terms of t?
  10. Sep 12, 2007 #9

    Doc Al

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    Staff: Mentor

    Right. (That's t in terms of V0, which is what you want.) When you use that to eliminate t in your equation for vertical motion, the only unknown will be V0.
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