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Ball rolled off top of stairs - which stair does it hit?

  1. Sep 28, 2012 #1
    1. The problem statement, all variables and given/known data
    A ball is rolled horizontally off the top of a staircase at 1.52 m/s. The dimensions of the steps are 20.3cm by 20.3cm in hight and width, respectively. Which step does the ball hit first?


    2. Relevant equations
    Vector kinematics equations


    3. The attempt at a solution
    At first I tried to make the stairs the slope of a line by using slope 20.3 anywhere on the Cartesian plane and then selected a start point where the projectile motion would begin.

    I tried to find out an equation for the projectile motion because it's parabolic and then find an intersection point between the parabola and a point on the line, then I'd divide that number accordingly with 20.3 to find which "stair" it hits first.

    Unfortunately I'm not sure how to transform a vector form into a quadratic equation. Is there any other way to do this?
     
  2. jcsd
  3. Sep 28, 2012 #2

    SammyS

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    Why did you choose a slope of 20.3 ?

    A slope of -1 is the appropriate slope.
     
  4. Sep 28, 2012 #3
    Remember that as the ball rolls off the edge the horizontal and vertical components of motion are independent of each other. The next question for you to solve then is this: In the time it will take the ball to fall 20.3 cm, how far will it have traveled in the horizontal direction at 1.52 m/s? If the distance is greater than 20.3 cm, then the ball has cleared the first step.
     
  5. Sep 28, 2012 #4
    Sorry, that's what I meant

    (Assuming the ball starts at height 0 and going down is negative velocity, cm is converted to meters)

    0.203 = -(1.52sin0)t - 4.9t^2
    4.9t^2 + 0.203 = 0
    t = 0.204 s

    x - x_0 = 1.52*0.204 + 0(constant velocity)
    So 0.310 meters in the horizontal direction, which would mean it's past the first step (since the first step is a horizontal distance of 0.203).

    Do I have to keep repeating this until I find the step? Is there a shorter method besides guess and check every additional step?
     
  6. Sep 28, 2012 #5

    Bandersnatch

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    Your initial approach, with finding intersection points, was correct.

    Take the position equation, [itex]r = \frac{1}{2}at^{2} + v_{0}t + r_{0}[/itex] and split it into two equations, one for the position on the x axis, one for the y.
    Now you want to find out at what time does the horizontal distance travelled equal the vertical distance travelled(downwards). x=-y, or y=-x.
    You've got now three equations with three variables. You can solve them for t.
    Having done that, it should be easy to find numerical value of the horizontal distance, and divide it by the width of a step.
     
  7. Sep 28, 2012 #6

    SammyS

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    For the horizontal direction:
    [itex]x=x_0+{v_0}_x\,t[/itex]
    [itex]=0+1.52\,t\ .[/itex]

    In the vertical direction:
    [itex]y=y_0+{v_0}_y\,t-\left(\frac{1}{2}\right)g\,t^2[/itex]
    [itex]=0+(0)\,t-4.9\,t^2\ .[/itex]

    and
    [itex]v_y={v_0}_y-g\,t[/itex]
    [itex]
    =0-9.8\,t\ .[/itex]​
     
  8. Sep 28, 2012 #7

    Borek

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    That's all assuming ball is a point :tongue2:
     
  9. Sep 28, 2012 #8
    How can you solve for t? Normally I'd use the y-component to find the time, but the y-component falls forever because there is no "bottom" to the stairs, therefore the time is infinite in the negative y direction (falling down).
     
  10. Sep 28, 2012 #9
    You find the path equation for the ball.....(1, eliminating time)
    Then find an equation where points on the equation are the end of the landings.....(2)

    Find the value of y at the point of intersection of those above equations.
    Finallly divide y with the height of the step.
     
  11. Sep 29, 2012 #10

    SammyS

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    Where does the ball cross the line, y = -x ?

    That tells you which step it lands on.

    How far does the ball fall (vertically) to land on that step?
     
  12. Sep 29, 2012 #11
    So for the position coordinates (x,y), we have (1.52t, -4.9t^2) which would mean -4.9t^2 = 1.52t and so the quadratic formula is 0 = 4.9t^2 + 1.52t

    Do I even need to solve for time? Can't I just make this into a qudratic formula in terms of t and then find the intersection points with the equation y = -t instead of with respect to x?

    Edit: Oh, never mind...the quadratic formula doesn't even make sense physically

    So how do I create a quadratic formula with respect to x?
     
    Last edited: Sep 29, 2012
  13. Sep 29, 2012 #12
    You have parametric equation of x and y.
    So what is y in term of x?
     
  14. Sep 29, 2012 #13

    Bandersnatch

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    Can't you just solve it? It'll give you the two moments in time when the ball trajectory intersect the y=-x slope. Then you just plug the result into the horizontal component of the position equation and voila!
     
  15. Sep 29, 2012 #14
    Admittedly I'm not that familiar with parametric equations.

    Given:

    y = -4.9t^2
    x = 1.52t

    Can I say that y = -4.9(x / 1.52)^2 ?

    [and then solve for -x = -4.9(x / 1.52)^2, which results in x = 0 and 0.47 as solutions]
    [plugging into the equation I get y = -0.47]
    [-0.47 / 0.204 = -2.30 which would indicate it's cleared the 2nd step and about 1/3 way into the 3rd step?]
     
  16. Sep 29, 2012 #15
    Yes.
    Since height and length are equal, you can use x value to calculate the steps which anyway you got the same answer.
     
    Last edited: Sep 29, 2012
  17. Sep 30, 2012 #16
    We know that if X>Y, making Y in the downward direction positive, then the ball clears the first step, but the same is true for the remaining steps. The point at which X=Y will tell us which step the ball landed on. X=vt and Y=1/2*a*t^2. You now need to solve for t. Knowing t will then give you a value for X. If 0≤X≤20.3 then it will land on the first step. If 20.3<X≤40.6, then it will land on the second step and so on.
     
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