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A ball is tossed from an upper-story window of a building. The ball is given an initial velocity of 8m/s at an angle of 20 below the horizontal. It strikes the ground 3.00s later. The horizontal distance at which the balls strike the ground from teh base of teh building and the height from which the ball was thrown are respectively given by:

Xf = Xi + Vxi*t

Change of X = Vxi*T

Xf = 8cos(20)(3)

Xf = 22.5 m

and Yi

Y = Vyi - 1/2g*t^2

0 = Yi + 8sin(20) - 1/2(9.81)(3)^2

Yi= 36m

are my answers right?

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# Homework Help: Motion in 2 dimension

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