A playground is on the flat roof of a city school, 5.5 m above the street below

Question

A playground is on the flat roof of a city school, 5.5 m above the street below (see figure). The vertical wall of the building is h = 6.90 m high, forming a 1.4-m-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; above the horizontal at a point d = 24.0 m from the base of the building wall. The ball takes 2.20 s 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 horizontal distance from the wall to the point on the roof where the ball lands.

Explanation / Answer

probably easiest with the trajectory eqn: y = h + x·tanT - g·x² / (2v²·cos²T) h = 0 x = 24 m T = 53º v = 18.18 m/s Plug & chug... y = 8.27 m So the ball clears the top of the wall by 8.27 - 6.1 = 2.17 m ? Employ the same eqn, but now x is unknown and y = 4.8m: y = h + x·tanT - g·x² / (2v²·cos²T) 4.8 = 0 + x•1.33 - x²•0.04 quadratic: a = 0.04 b = -1.33 c = 4.8 x = 4.1m, 29.1m The two values for x represent the two distances from the passerby for which the height of the ball is 4.8m. Clearly the second one is the one that interests us (as it has cleared the wall). Therefore the horizontal distance from the wall to the point on the roof where the ball lands is 29.1 - 24 = 5.1 m

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