Physics
posted by Crystal .
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^2Yo^2)/2g=(03.5^2)/19.6 = 0.625 m.
e. Y = Yo + g*t = 0 @ max. ht.
Tr = (VVo)/g = (03.5)/9.8=0.357 s =
Rise time.
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.
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