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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 the base of the 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?
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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