Posted by emanuel on .
From the window of a building, a ball is tossed from a height y 0 above the ground with an initial velocity of 10.00 m/s and angle of 25.0° below the horizontal. It strikes the ground 3.60 s later.
If the base of the building is taken to be the origin of the coordinates, with upward the positive y- direction, determine:
a) How far horizontally from the base of the building does the ball strike the ground (5 points)
b) the height from which the ball was thrown (5 points)
s) the velocity with which it hits the ground below (5 points)
Vo = 10m/s[-25o]
Xo = 10*cos(-25) = 9.063 m/s
Yo = 10*sin(-25) = -4.226 m/s.
a. D=Xo * Tf = 9.063m/s * 3.60s=32.63 m.
b. h = Yo*t + 0.5g*t^2
h = -4.226*3.6 + 4.9*3.6^2 = 48.29 m.
c. Y^2 = Yo^2 + 2g*h
Y^2 = -4.226^2 + 19.6*48.29 = 964.3 m.
Y = 31.05 m/s.