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2D kinematics question

  1. Oct 30, 2009 #1
    1. The problem statement, all variables and given/known data

    A softball is hit over a third baseman's head with some speed v-sub0 at an angle theta above the horizontal. Immediately after the ball is hit, the third baseman turns around and begins to run at a constant velocity V = 7.00m/s. He catches the ball t=2.00sec later at the same height at which it left the bat. The third baseman was originally standing L=18.0 m from the location at which the ball was hit. Find v-sub0 Use g=9.81 m/s^2 for the magnitude of the acceleration due to gravity. also find theta.

    2. Relevant equations

    x =( initial velocity times cos theta)* time
    y = initial velocity times sin theta)* time - .5GT^2

    3. The attempt at a solution

    i dont understand how to solve an equation with 4 variables and i only have values for 2 of them. i have time = 2 seconds, and distance = 32 meters. how do i get an angle and an initial velocity?
     
  2. jcsd
  3. Oct 30, 2009 #2

    Doc Al

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    Staff: Mentor

    There are two equations and two unknowns. Hint: What's y, measured from the starting height?
     
  4. Oct 30, 2009 #3
    is it 9.8 meters?
     
  5. Oct 30, 2009 #4

    Doc Al

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    No. How does the initial height compare to the final height?
     
  6. Oct 30, 2009 #5
    it seems like you just use the dude running at 7 m/s for 2 seconds to get the total distance (18m + 14m) but i guess thats not the right way to go about it?


    i really wish my physics prof spoke english!
     
  7. Oct 30, 2009 #6
    they are the same.
     
  8. Oct 30, 2009 #7

    Doc Al

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    No, that's perfectly correct. That gives you the value for x you'll need in the first equation.

    What about y?
     
  9. Oct 30, 2009 #8

    Doc Al

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    Yes! So, if you measure from the starting point, what's y? (What's the change in height?)
     
  10. Oct 30, 2009 #9
    zero?
     
  11. Oct 30, 2009 #10

    Doc Al

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    Yes, y = 0 when t = 2.
     
  12. Oct 30, 2009 #11
    so i have:
    32=(initial velocity*cos theta)*time
    and
    0= (initial velocity*sin theta)*time - .5GT^2
    right?
    but now what?
     
  13. Oct 30, 2009 #12
    wait, i have 16 = initial velocity * cos theta
    and 9.8 = initial velocity * sin theta (or is it 4.9/2 = initial Velocity * sin theta ?)

    (i think)
    now what?
     
  14. Oct 30, 2009 #13

    Doc Al

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    Play around with those equations and see if you can isolate one of the variables. There are several ways to go. (Try division.)
     
  15. Oct 30, 2009 #14
    OK, maybe 16/cos theta = 9.8/sin theta?

    im so lost....
     
  16. Oct 30, 2009 #15
    am i warm?
     
  17. Oct 30, 2009 #16
    is it an easy Trigonometric solution? im in Trig/precalc algebra in one class and we just started the Trig half, so maybe im not familiar with a function to equate Sin/Cos?
     
  18. Oct 30, 2009 #17

    Doc Al

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    You're not that lost. Multiply both sides by sin theta. (What other trig function appears?)
     
  19. Oct 30, 2009 #18
    does Tan theta = 9.8/16 ?
    am i on the right track?
     
  20. Oct 30, 2009 #19
    holy mackeral!!! thanks Doc Al, i really appreciate the help!
     
  21. Oct 30, 2009 #20

    Doc Al

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    You got it. :approve:
     
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