posted by Crystal on .
A soccer ball iskicked with an intial horizontal speed of 6 m/s and initial vertical speed of 3.5 m/s. Assume that the projection and landing height are the same, and neglect air resistance. Calculate the following
a) the ball's projection speed and angle
b)the ball's horizontal speed at .5 s into its flight
c) The ball's vertical speed at .5 s into its flight
d) the ball's height at its apex
e) the ball's flight time
f)the ball's flight distance
Xo = 6 m/s.
Yo = 3.5 m/s.
a. tanA = Yo/Xo = 3.5/6 = 0.58333
A = 30.3o
Vo=Xo/cosA = 6/cos30.3 = 6.95m/s[30.3o]
b. X = Xo = 6 m/s. It does not change.
c. Y = Yo + g*t
Y = 3.5 - 9.8*0.5 = -1.4 m/s. The negative sign means the ball is falling.
d. h=(Y^2-Yo^2)/2g=(0-3.5^2)/-19.6 = 0.625 m.
e. Y = Yo + g*t = 0 @ max. ht.
Tr = (V-Vo)/g = (0-3.5)/-9.8=0.357 s =
Tf = Tr = 0.357 s. = Fall time.
T = Tr + Tf = 0.357 + 0.367 = 0.714 s.
f. D = Xo * t = 6m/s * 0.714s = 4.29 m.
= 0.357 = 0.714 s.