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Uniformly Accelerated Projectile Motion

  1. Jul 12, 2008 #1

    Here's my probelm:

    A baseball is thrown with an initial velocity of 100 m/s at an angle of 30° above the horizontal. How far from the throwing point will the baseball attain its original level?

    The attempt at a solution:
    vix = 100 cos 30° => 86.6 m/s
    viy = 100 cos 30° => 50 m/s

    To find the distance we need to use [tex]v = \frac{d}{t} [/tex] = d = vt
    Now in order to find the time we can use;
    [tex]y = v_{iy}t + \frac{1}{2}a_{y}t^2[/tex]

    [tex]y = 0[/tex] since the ball is coming back to its original height.
    = [tex]0 = (50 m/s) . t + \frac{1}{2} (-9.8 m/s^2 ) t^2[/tex]

    Am I right.....? I can't exactly see how to obtain the time out of this…


    Last edited: Jul 12, 2008
  2. jcsd
  3. Jul 12, 2008 #2
    That's right -- from there, you can tell that there are two values for t when y = 0 -- t = 0 means the moment the ball was launched, and the other t will signify when the ball hits the ground again. Of course, this is given the fact that the ground is the point of reference and not some elevated height. Solve via algebra for the other t and you're done.
  4. Jul 12, 2008 #3
    Thank you,

    But I didn't quite understand the last part, you mean we take the first t = 0 ?
    [tex]0 = (50 m/s) . 0 + \frac{1}{2} (-9.8 m/s^2 ) t^2[/tex]

    And then we sove for t² ?

    .....and once we find time, we can use [tex]x = v_{x}t[/tex] to find the distance the ball traveled until it attained its original level.
  5. Jul 12, 2008 #4
    No, he means that one of the answers to that equation will be t = 0 (because at t=0, the ball was in fact at the original level). You need to solve for the other solution to that equation, just put one of the terms on the other side of the "=" and cancel out one of the [tex]t[/tex]s
  6. Jul 12, 2008 #5

    Which one of the terms t ot t^2? It's kind of confusing, that's where I'm stuck...

    [tex](50 m/s) t + \frac{1}{2} (-9.8 m/s^2 ) t^2 = 0[/tex]

    If we take t to the other side we get;

    [tex](50 m/s) + \frac{1}{2} (-9.8 m/s^2 ) t^2 = \frac{1}{t}[/tex]

  7. Jul 12, 2008 #6
    ok, if you want to find the distance the object will obtain when it reaches the horizon once more, you do this:

    A) use a=(vf-vi)/t
    this equation is changed to t=(vf-vi)/a
    t=5.09s **** that is the time from the horizon to the apex****
    total time = 5.09s*2 = 10.18s

    B) now you can use d=v*t
    d=86.6m/s * 10.18s
  8. Jul 12, 2008 #7
    But I am rather curious as how to obtain time from: [tex](50 m/s) t + \frac{1}{2} (-9.8 m/s^2 ) t^2 = 0[/tex]


    Last edited: Jul 12, 2008
  9. Jul 12, 2008 #8
    well, if you really want to do math other than physics to solve this problem using d=vi*t+0.5*a*t^2 all you have to do is plugin for the equation. you do not need to solve for d. Because you want to know the time when the ball hits the floor, all you have to do is find when the distance is going to be 0. that really happens 2 times: before the ball leaves and when it lands *****bouncing doesnt count*****
    therefore, you use the quadratic formula on this d=50*t+9.81/2*t^2. d is like y-axis do that has to be 0 for the answer and t is the x-axis.
    the answers are x=[0 and 10.1937] when y =0
    you get it?
  10. Jul 12, 2008 #9
    Oh, that makes perfect sense now. Thanks. :smile::smile:
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