Posted by <3 on .
From the window of a building, a ball is tossed from a height y0 above the ground with an initial velocity of 9.40 m/s and angle of 23.0° below the horizontal. It strikes the ground 6.00 s later.
Initial velocity components:
vx = 8.65 m/s
vy = 3.67 m/s
(c) Find the equations for the x and y components of the position as functions of time.
x = 8.65t
y = ?
(d) How far horizontally from the base of the building does the ball strike the ground?
51.9 m
(e) Find the height from which the ball was thrown.
(f) How long does it take the ball to reach a point 10.0 m below the level of launching?
I need help on (c), (e), and (f).

Physics 
Henry,
Vo = 9.4m/s[23o]
Xo = 9.4*cos23 = 8.65 m/s.
Yo = 9.4*sin23 = 3.67 m/s.
c. X = Xo * t = 8.65t
Y = Yo*t + 0.5g*t^2=3.67t+4.9t^2
e. h = 3.67t + 4.9t^2
h = Y = 3.67*6 + 4.9*6^2 = 198.4 m.
Above gnd.
f. 3.67t + 4.9t^2 = 10 m.
4.9t^2 + 3,67t _ 10 = 0
t = 1.10 s. (Use Quad. Formula).