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.(adsbygoogle = window.adsbygoogle || []).push({});

(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 xiand yiin equations. I am...just confused. If using yf= yi+ vyit + .5a(t^2), the new yiwould be the old yf(which is 7.263...m). The new vyiis 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+ aytmaywork, 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

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# Homework Help: Playground ball trajectory.

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