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Help with projectile motion problem! Solution?

  1. Feb 28, 2008 #1
    1. The problem statement, all variables and given/known data
    A home run is hit in such a way that the baseball just clears a wall 21m high, located 130m from home plate. The ball is hit at such an angle of 35degrees to the horizontal, and air resistance is negligible. Find initial speed, the time it takes the ball to reach the wall, and the velocity components and the speed of the ball when it reaches the wall. (Assume that the ball is hit at a height of 1m above the ground. Suggestions or solutions please? I'm stumped!


    2. Relevant equations
    x=vo X t + .5 X a X t^squared


    3. The attempt at a solution

    ???
     
  2. jcsd
  3. Feb 28, 2008 #2

    Dick

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    x(t)=v0_x*t (there is no acceleration in the x direction).
    y(t)=y0+v0_y*t+(1/2)*g*t^2 (acceleration in the y direction is g downward).
    v0_x and v0_y are related to the initial speed v0 by trig functions of the angle, right? I don't think they would give this question to a complete beginner. You must be able to do a better job of at least starting to work out the question. If t=0 is when the ball is hit. What is y0?
     
  4. Feb 29, 2008 #3
    Help?

    Ok... well, a^y = -9.8 m/s^2
    theta = 35 degrees
    y = 21 - 1 = 20 m, because of the ball being hit 1 m above the ground
    x = 130 m

    But you can't solve for the initial velocity or the horizontal/vertical components... how do i solve for the initial velocity? Or the time?

    Vo = x / t
     
  5. Feb 29, 2008 #4

    Dick

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    v0_x=v0*cos(theta), v0_y=v0*sin(theta). Haven't you ever seen anything like that before? Split the balls motion into x and y components!
     
  6. Feb 29, 2008 #5
    ok... so the initial velocity in the x direction is Vox = Vo X cos (theta).
    since you don't have the initial velocity it cancels out the cos (theta). Correct?
    Same goes with y initial velocity.

    Vo X .819 = Vox
    Vo X .574 = Voy
     
  7. Feb 29, 2008 #6

    Dick

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    Nothing is cancelling anything out. The numerical value of cos(theta)=0.819. That's ok. You don't have the initial velocity so you have to leave it in the equations as an unknown until you can solve for it in the end. Next, use the equation describing the x component of the motion to find the time to reach the fence. Your answer for t will have the unknown v0 in it.
     
  8. Feb 29, 2008 #7
    ok well I'm confused... I have the answers in the back of the book to check my answers but that's not the problem. I want to know how to get those answers.

    If the equation x = Vo x t, then 130 = Vo x t. I still have 2 unknowns in this problem. The initial velocity and the time. I can't solve for 1 if I don't have the other. Unless I am missing something here and completely oblivious to what is in front of me.

    Here's the thing, if the answer is 3.8s and to find the initial velocity (Vo), I would take 130/3.8 = 34.2 m/s. But that is not the correct answer.
     
  9. Feb 29, 2008 #8

    Dick

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    t=(130m)/(v0_x). That's the time when it reaches the fence. Yes, you still have two unknowns in the problem. That's ok, because we aren't done. Now write down the equation for the y motion.
     
  10. Feb 29, 2008 #9
    y = Vyo X t - 1/2 a X t^2

    20 = Vyo X t - .5(-10m/s)t^2

    Unknown again is initial velocity and time
     
  11. Feb 29, 2008 #10
    Well, I'll try to work on this more this weekend. I'll be out of town until Sunday afternoon, so I'll take a look at it then. Thanks for all of your help. I'm determined to figure this problem out!
     
  12. Feb 29, 2008 #11

    Dick

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    Ok, so t=130m/(v0_x) is when it reaches the fence. Put that value of t into the y equation. Now there is only one unknown, v0.
     
  13. Mar 1, 2008 #12
    y = Vyo X t - 1/2 a X t^2

    20 = Vyo X 130m/(v0_x) - .5(-10m/s)130m/(v0_x)^2
     
  14. Mar 1, 2008 #13

    Dick

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    I wish you'd try a little harder. You got that stuff you wrote before. And t^2=(130m/v0x)^2. You could finish this anytime you want.
     
  15. Jul 25, 2010 #14
    dx=(Vix)(t)
    130=(Vcos35)(-2(Vsin35)/-9.8)
    130=V^2(cos35)(0.117)
    V^2=130/0.096
    V=36.82 m/s

    t=(-2(Viy)/g)
    t=(-2(Vsin35)/-9.8) or t=(-2(21.12 m/s)/-9.8)
    t=4.31 s

    Vy= Vsin35 = 21.12 m/s
    Vx= Vcos35 = 30.16 m/s

    I hope this helped!
     
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