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Projectile Motion

  1. Dec 25, 2006 #1
    A ball rolls off a stairway with a horizontal velocity of 5.0 ft/sec. The steps are 8.0 in high and 8.0 in wide. Which step will the ball land on first?

    I was thinking we could find the time it takes for the ball to go 8in in the X direction(using our motion equations), then plug that in for the Y location and see where we would be. After that I am not sure, I would know if we had the height of the stairway.
  2. jcsd
  3. Dec 25, 2006 #2
    hmm... what an interesting problem.

    How about in general, a ball being projected from a stair; given the height and width of each step, find the step which the ball will land on first.

    I think the solution is quite simple. imagine drawing a line, l, by connecting each "corner" of the steps. For the ball to land on some step, the ball must pass the portion of line l above the step first. Due to the negative concavity of the parabolic projectile motion of the ball, if the ball passes the portion of l above the nth step, then it must land on step n.

    So, if you find the intersection between line l and the path, you can find the solution.
  4. Dec 26, 2006 #3
    Thanks tim_lou. I am aware that I could set tis up graphically and find the intersection of L. However, how about expressing it in mathematical terms?
  5. Dec 26, 2006 #4
    Are you aware that you are indeed given the height of each step?
    I agree with you that you should be looking at time first. (But I'd focus on the vertical rather than the horizontal). What equation do you know that relates initial y-velocity with time?
    Last edited: Dec 26, 2006
  6. Dec 26, 2006 #5
    I would say I would have to use y=y(knot)+Vt-1/2gt^2. Using the refernce point where the ball is y(knot)=0. We are not given the intial velocity on the verticle and Vt can be stratched out leaving us with y=-1/2gt^2. We set Y=8 to see how long it takes for the ball to travel 8in, right? Oh yah, I would either have to convert my g to in/sec, or I could conver my in of the step into FT to have compatible units.
    Last edited: Dec 26, 2006
  7. Dec 26, 2006 #6
    just put t in terms of x.
    [tex]x=v_0\cos\theta t[/tex]
    *to be convenient, choose the right coordinates so that the initial x displacement as zero.
    then, find the intersection between the function of the parabola and the line.

    find the y value of the intersection, that might be some messy algebra but it is doable. In general, you can express n in terms of initial conditions using floor or ceiling function.
    Last edited: Dec 26, 2006
  8. Dec 27, 2006 #7
    I like the y(knot) notation, but I believe the common spelling is y(nought), nought being 0. Or you could just write y_0.

    It looks to me as if we have been given the initial y-velocity--it's 0. (so you're calculation holds even if your reasoning doesn't)

    Looks good to me! Also, if the ball will travel, say 1 1/2 steps horizontally in the time it takes to fall one step vertically, you'll need to do more calculating (because the ball will now have to fall at least 16" before landing.) Actually, at this point there should be an easier method than what I'm suggesting.
  9. Dec 28, 2006 #8
    But since the ball has a curved path should I be using [tex]x=v_0\cos\theta t[/tex] or [tex]y=v_0\sin\theta t-\frac{1}{2}gt^2[/tex]
    Last edited: Dec 28, 2006
  10. Dec 28, 2006 #9

    Doc Al

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    Staff: Mentor

    You need to use equations for both vertical and horizontal motion. But you can simplify things greatly, since the initial velocity is horizontal.

    You had said:
    You are given the initial vertical component of velocity! :wink:
    Last edited: Dec 28, 2006
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