A playground is on the flat roof of a city school, 5.3 m above the street below

Question

A playground is on the flat roof of a city school, 5.3 m above the street below (see figure). The vertical wall of the building is h = 6.70 m high, to form 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.


http://www.webassign.net/sercp9/3-p-034.gif


(a) Find the speed at which the ball was launched. m/s


(b) Find the vertical distance by which the ball clears the wall. m


(c) Find the horizontal distance from the wall to the point on the roof where the ball lands. m

Explanation / Answer

given that the ball takes 2.2 s to reach the vertical point above thee wall => hotizontal distance travelled in that time is 24 m so 24 = ux * 2.2 => ux = 10.9 m/s and u cos 53 = ux so initial velocity u = ux/(cos 53) =18.11 m/s uy = 18.11 sin 53=14.46 m/s => the height reached by the ball in 2.2 secs is h = uy * t - 0.5 * g *t^2 h =8.1 m so the vertical distance by which the ball clears the wall = 8.1-6.7= 1.4 m time taken by the ball to reach 5.3 m height is 5.3 = 14.46 *t - 0.5 *9.8 *t^2 =>t=2.52 s horizontal distance travelled in this time = 2.52 * 10.9 = 27.47 m so horizontal distance from the wall to the point on the roof where the ball lands = 27.47-24 = 3.47 m

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