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Homework Help: Hard 2d motion problem

  1. Sep 22, 2009 #1
    1. The problem statement, all variables and given/known data

    A ball player hits a home run, and the baseball
    just clears a wall 8.00 m high located 122.5 m
    from home plate. The ball is hit at an angle
    of 39.9◦ to the horizontal, and air resistance is
    negligible. Assume the ball is hit at a height
    of 1.2 m above the ground.
    The acceleration of gravity is 9.81 m/s2 .
    a) What is the initial speed of the ball?
    Answer in units of m/s.

    2. Relevant equations

    y= yo+sinLaTeX Code: \\theta *t-1/2gt^2

    1/2gt/yo= voy

    R= Voy*t=VosinLaTeX Code: \\theta *t

    t= LaTeX Code: \\sqrt{} (2y/g)^2

    3. The attempt at a solution

    I plugged in my known variables but couldn't get anything

    Please help
  2. jcsd
  3. Sep 22, 2009 #2
    According to your equations, it looks like you began in the right direction, but then steered off course a bit.

    One thing to note is you subtracted (1/2)*g*t2. Be careful that you don't put -9.8m/s2 as g, since you prematurely subtracted it.

    You began with

    then you substituted v0y with sin(39.9). It should be v0y=v0*sin(39.9).

    I'm not sure what you did next. I would like to see how you derived (1/2)*g*t/y0=v0y.
    I'm pretty sure it's nonphysical, though. When I solved it, I pretended the ball started at a height of 0 (y0=0m) and traveled a height of 8m-1.2m (y=6.8m) so your equation would have a 0 in the denominator in my case.

    I'm not certain what you intend R to represent. You're correct in stating v0y*t=v0*sin(39.9)*t, which is actually the flaw you made earlier in substitution, leaving v0 out.

    t=sqrt(((2y)/g)2)? Well first the square root and the square cancel out becoming t=(2y)/g. I'm not sure how you derived this, and I'm pretty sure it's nonphysical. EDIT: Ah, I think you tried to find the initial velocity by substituting time from your original equation INTO your original equation. Well, again I'm pretty sure that you didn't rearrange the equation correctly, and you need a second equation with the same unknowns to do this type of substitution. What I think you did was, say x-y=1, thus x=1+y, thus 1+y-y=1, which of course 1=1, which says nothing about x or y.

    It looks like you understand that you'll have two unknowns (t and v0). I think you went back to your original equation and rearranged it after having attempted substituting for v0y. I would continue working with just substituting for your unknowns and leave rearranging your original equation at the end. You just need to create two equations with two unknowns. It appears you were going in the right direction, but made some serious algebra errors. Perhaps try starting with the x components first. That will be your second equation, and much simpler equation.

    By the way if there is an easier way to solve this problem, I'm sorry I don't see it. Otherwise this is a long way to get the answer. =)
    Last edited: Sep 22, 2009
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