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Two - dimensional motion problem

  1. May 29, 2006 #1
    hi i need some help with this problem...

    a home run is hit in such a way that the baseball just clears a wall 21 m high, located 130 m from home plate. The ball is hit at an angle of 35 degrees to the horizontal, and air resistance is negligible.

    what's the initial speed of ball?

    this is what i did

    x=Vxot Vxo=(cos 45)(Vo) and X=130(horizontal distance)

    now i try to find t...so i

    change in Y=Vy0t - .5g(t^2) so with this i get 21=Sin 45(Yo)t-4.9(t^2)

    i simplify that to 20=t(.57v - 4.9t)
    i get t=20 or t=.57v/4.9 and i plug the latter into the first equation but i dont get the right answer which is 42 m/s

    im thinking that i messed up on my algebra but i don't know.

    I would appreciate if someone could help me...thank you
  2. jcsd
  3. May 30, 2006 #2


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    Homework Helper

    Your approach is ok, 2 equations and 2 unknowns (Vo and t), and by the way the angle your problem statement uses is 35 degrees, not 45.

    The equations:

    [tex] x = v(0) \cos \theta t [/tex]

    [tex] y = v(0) \sin \theta - \frac{1}{2} gt^2 [/tex]
  4. May 30, 2006 #3
    oh thats as a typo...im not getting right answer with this approach
  5. May 30, 2006 #4
    Remember that you're not equating that ^ to zero. Also, the y = 21 as per the problem.
  6. May 30, 2006 #5
    oh okay...so i got .57Vot - 4.9t^2 = 21 with that i get t=21 or
    t=(21 -.57Vo)/(-4.9)

    so with this i plug the latter into the first equation:

    130=Vo(cos A)t t=(21 -.57Vo)/(-4.9)

    so i get

    130=Vo (cos 35)(21 -.57Vo)/(-4.9)

    i simplify this and end up getting 13 for Vo which wrong....what am i doing wrong?
  7. May 31, 2006 #6
    So you're essentially saying is that when
    [tex](0.57v_o - 4.9t)t = 21[/tex], t (can be) = 21.

    From this, we can see that [tex](0.57v_o - 4.9t)[/tex] should be equal to 1. And from the aforementioned value of t, we can solve for [tex]v_o[/tex], without the need for the first equation ([tex]x = v_o \cos\theta_o t[/tex]). Seems too simple, doesn't it? :wink:
  8. Jun 1, 2006 #7
    i did that i and ended up with Vo = 182.28 but the answer is 42 :(
  9. Jun 1, 2006 #8
    Shouldn't the equation for y(t) be

    [tex] y = v(0) \sin \theta t - \frac{1}{2} gt^2 [/tex]

  10. Jun 1, 2006 #9
    I'm sorry, I should've been a bit more clear. I was trying to show you that it wasn't the correct way. You solve the quadratic equation 0.57vt - 4.9t² - 21 = 0 and find the appropriate value of t.
  11. Jun 1, 2006 #10
    im not sure i know how to do that with .57vt - 4.9t^2- 21 = 0

    mann i feel stupid right now
  12. Jun 1, 2006 #11
    The roots of a quadratic equation [tex]ax^2 + bx + c = 0 (a \neq 0)[/tex] are [tex] x = \frac{-b+\sqrt{b^2 - 4ac}}{2a}[/tex] and [tex]x = \frac{-b-\sqrt{b^2 - 4ac}}{2a}[/tex]. Can you solve the equation now?
  13. Jun 1, 2006 #12
    yeah, so do i make B= .57vt?

    if so, then ill get...-.57+or- Square root of(.3249v^2)-411.6) / (-9.8)

    ..i wouldnt know what to do from here....
  14. Jun 1, 2006 #13
    You must take care to check the sign of the terms. If b = 0.57vt, then a should be -4.9, or if you find it easier, you can rewrite it such a is positive. .
  15. Jun 1, 2006 #14
    i just need help simplifying the problem and solving for t....can somebody plz just show me how one would get 42 as the initial velocity...
    thank you
  16. Jun 1, 2006 #15


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    Gold Member

    Okay, as you are clearly struggling with this question I will walk you through it step by step. If we split the initial velocity(v) into horizontal (x) and vertical (y) components;

    [tex]v_{x} = v\cos 35[/tex]

    [tex]v_{y} = v\sin 35[/tex]

    Now, you know that to clear the wall the vertical displacement must be 21m and the horizontal displacement must be 130m.

    [tex]s_{x} = 130[/tex]

    [tex]s_{y} = 20[/tex]

    Now in the vertical direction the acceleration is going to be that due to gravity. However, as we are ignoring air resistance, there is no acceleration in the horizontal plane. Using the kinematic equation [itex]s = ut + \frac{1}{2}at^{2}[/tex] We can now write two equations, one for horizontal motion and one for vertical motion;


    As there is no acceleration in the horizontal direction the equation becomes;

    [tex]s = ut[/tex]

    Substituting the numbers in;

    [tex]130 = v\cos 35 t[/tex]

    We can then rearrange this into a function for time;

    [tex]t = \frac{130}{v\cos 35}[/tex]

    Let this equation be (1)


    Substituing the values in gives;

    [tex]21 = v\sin 35 t + -4.905 t^2[/tex]

    Now if you substitute equation (1) into the above equation you can eliminate t and solve for v. You will obtain a quadratic equation in terms of v, you can solve for v using the quadratic equation. You would also do well to note the trig identity;

    [tex]\tan\theta = \frac{\sin\theta}{\cos\theta}[/tex]

    Do you follow?

    Last edited: Jun 1, 2006
  17. Jun 2, 2006 #16
    thanks a lot everybody...i really appreciate the help
  18. Oct 2, 2007 #17
    i don't get how u still got 42m/s.. i have this same problem as well.. plese.. explain more with detail.. i'm sorry.. i'm so stupid...T-T
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