10:05 PM webassign.net Need Help?Read It Need Help? Read It +-/3 points SerPSET9

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10:05 PM webassign.net Need Help?Read It Need Help? Read It +-/3 points SerPSET9 4.P.020 My Notes Ask Your Teacl A ball is tossed from an upper-story window of a building. The ball is given an initial velocity of 8.10 m/s at an angle of 18.0° below the horizontal. It strikes the ground 6.00 s later (a) How far horizontally from the base of the building does the ball strike the ground? (b) Find the height from which the ball was thrown. (c) How long does it take the ball to reach a point 10.0 m below the level of launching? Need Help? Read ItWatch It Submit Answer Save Progress . Practice Another Version + -/3 points SerPSET94-P.025 My Notes Ask Your Teac! A playground is on the flat roof of a city school, 5.7 m above the street below (see figure). The vertical wall of the building is h = 7.20 m high, forming a 1.5-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.

Explanation / Answer

(a) x = u cos 18 * t
= 8.10*cos 18 * 6s = 32.09 m

(b) h = u * sin 18* t + 1/2 * g * t^2
= 8.10 sin18 * 6 + 1/2 * 9.8 *62 = 139.91 m

(c) y = u * sin 18* t + 1/2 * g * t^2
10 = 8.10 * sin 18*t + 4.9t^2
--> 4.9t2 - 6.08t - 10 = 0

by sloving the above equation we get
---> t = 2.177 s

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