A playground is on the flat roof of a city school, 5.1 m above the street below

Question

A playground is on the flat roof of a city school, 5.1 m above the street below (see figure). The vertical wall of the building is h = 6.40 m high, forming a 1.3-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. find speed at which ball was launched?, find vertical distance by which ball clears the wall?,

Explanation / Answer

a.) To find the speed at which the ball was launched: D = v * t 24 m = (V*cos(53 degrees)) * 2.2 s where Vcos(angle) = horizontal speed V = 18.13 m/s b.) At that initial speed, the initial vertical component is: Vy = V*sin(angle) Vy = 18.13*sin(53 degrees) Vy = 14.48 m/s To get vertical distance, use: d = Vy *t + (1/2) * a * t^2 d = 14.48 * 2.2 + (1/2) * (-9.8) * 2.2^2 d = 8.14 m So the ball clears by 8.14m - 6.4 m = 1.74m

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