physics
posted by Teddy on .
A)A ball is tossed from an upperstory window of a building. The ball is given an initial velocity of 8.15 m/s at an angle of 19.0o below the horizontal. It strikes the ground 2.50 s later. How far horizontally from the base of the building does the ball strike the ground?
B)Calculate the height from which the ball was thrown.
C)How long does it take the ball to reach a point 12.0 m below the level of launching?

Vo = 8.15m/s[19o]
Xo = 8.15*cos(19) = 7.71 m/s.
Yo = 8.15*sin(19) = 2.65 m/s.
A.Dx=Xo * Tf = 7.71m/s * 2.50s=19.26 m/s
B. h = Yo*T + 0.5g*T^2
h = 2.65*2.5 + 4.9*2.5^2 = 24 m.
C. 2.65*T + 4.9T^2 = 2412 = 12
4.9T^2  2.65T  12 = 0
T = 1.86 s.