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?
(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).
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.
f. 3.67t + 4.9t^2 = 10 m.
4.9t^2 + 3,67t _- 10 = 0
t = 1.10 s. (Use Quad. Formula).