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Playground ball trajectory.

  1. Aug 21, 2007 #1
    A playground is on the flat roof of a city school, 6.00m above the street below. The vertical wall of the building is 7.00m high, to for a meter-high railing around the playground. A ball has fallen to the street below, and a passerby returns it by launching it at an angle of 53.0 degrees above the horizontal at a point 24.0 meters from the base of the building wall. The ball takes 2.20s to reach a point vertically above the wall. (a) Find the speed at which the ball was launched. (b) Find the vertical distance by which the ball clears the wall. (c) Find the distance from the wall to the point on the roof where the ball lands.

    (a) xf = xi + vxit + .5a(t^2)

    My thinking is that the horizontal distance the ball must travel is 24 feet to be in line with the wall. It is at that wall in 2.24 seconds. so, for that portion of the trajectory, the final x component is 24m, and t = 2.24s. Acceleration would be, of course, 0.

    24.0 = 0 + vxi(2.24) + .5(0)(t^2)
    vxi = 10.71428571 (will round later...as I still have more calculating to do)

    So, since vxi = vicosθ, 10.71428571 = vicos53.

    vi = 17.80...m/s

    (b) yf = yi + vyit + .5a(t^2)

    The initial y portion is at zero before the ball is thrown. I can find out the y portion of the initial velocity by taking the initial velocity of (a) and multiplying it by sine of 53 degrees. I get 14.218...m/s. Since I'm trying to find the y component that corresponds to x distance of 24 m, I use t = 2.24:

    yf = 0 + (14.218)(2.24) + .5(- 9.8)(2.24^2)

    = 7.2628...m

    Because the wall is 7.00 m, this ball, at time 2.24 seconds, clears the wall by .2628 m.

    (c) If my train of thought is correct on the last two parts, then that's good. Because I'm having trouble with (c). My first thought is using that point where t = 2.24 as an initial point, so the x and y components become xi and yi in equations. I am...just confused. If using yf = yi + vyit + .5a(t^2), the new yi would be the old yf (which is 7.263...m). The new vyi is the old vyf, which can be figured out by multiplying -9.8m/(s^2) by 2.24s. The time it takes from the wall to hitting the roof is tricky though, and that's what I cannot figure out. I suppose the vyf = vyi + ayt may work, if I assume the final velocity (once it hits the roof) is the negative of the original initial velocity (way back when the ball was thrown). But, that's the problem. The roof is an obstruction--something that disrupts the usual parabolic path. If I wanted the velocity upon roof-to-ball contact, I would need the time, and, well, that's what I also need. ...this is going in a circle around my head, and I probably need to bring another equation into the mix. XDDD
     
  2. jcsd
  3. Aug 21, 2007 #2
    a) Does it take 2.24 seconds to reach the wall, or 2.20? If you don't want to keep track of decimal places, just say that v_x = 24/2.20 and keep it that way. That will save you way more time and errors than saving all the decimals.

    b) Given that you used the correct time, this is right.

    c) You may be making this more complicated than it is. At how many times in the balls trajectory will it cross the 6m mark? What do these times mean, which one goes with what?
     
  4. Aug 22, 2007 #3
    But, even with the decimals, for (a), my logic is correct, right?

    As for (c), the ball does cross the level of the playground four times, but it's like two separate parabolas. 6.00m is the y component of the parabola...and on the return trip it is touched twice...so there would be some quadratic equation involved, which means I would use yf = yi + vyit + .5a(t^2), with final y = 6 and the initial y = 0? And...vyi would be 14.218m/s if (a) is correct, as well as a = -9.8...

    And, if I'm given two times, I would take the latter time, because it cannot hit on the first time (unless it executes a certain velocity to where it hits the building's very edge at its max, there is no way a ball can go past the building's edge to land on it without exceeding the building's level of...6.00 m).

    And, once I get this time, I can use xf = xi + vxit + .5a(t^2), with xi = 0, a = 0, t = the larger time from the quad. formula, and vxi = 10.714...m/s since this time, I'm using the entire trajectory of the ball coming back to the building.

    ...right?
     
  5. Aug 22, 2007 #4
    Ignore the vx component of the flight. Imagine the ball is being thrown straight up. It will go past the height of 6m, reach a maximum, and then it will fall back down and go past 6m again.

    Create a quadratic that will allow you to solve for the 2 values of t where the y is equal to 6. It's not that hard. One is less than 2.24 seconds, and the other is greater.

    Then, just define the point on the wall where the ball is above as x = 0, and the time the ball travels as (t2 - 2.24), it then becomes a simple linear problem. d = vt.
     
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