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20 July, 02:36

Ou are walking around your neighborhood and you see a child on top of a roof of a building kick a soccer ball. the soccer ball is kicked at 47° from the edge of the building with an initial velocity of 21 m/s and lands 57 meters away from the wall. how tall is the building that the child is standing on?

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  1. 20 July, 02:44
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    9.16 meters First, split the velocity into horizontal and vertical components. h = 21 cos (47°) = 21 * 0.744270977 = 15.62969052 m/s v = 21 sin (47°) = 21 * 0.731353702 = 15.35842773 m/s Now determine how many seconds the ball had to travel to reach 57 meters. T = 57 m / 15.62969052 m/s = 3.647 s height of the ball at time T is d = vT - 0.5AT^2 where v = initial velocity T = Time A = acceleration due to gravity (9.8m/s^2) Plug in the known values d = (15.35842773 m/s) (3.647 s) - 0.5 9.8 m/s^2 (3.647 s) ^2 d = 56.01219 m - 4.9 m/s^2 (13.30061s^2) d = 56.01219 m - 65.17298 m d = - 9.1608 m So the ball fell a total of 9.16 meters, which means that the building was 9.16 meters tall.
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