Physics
posted by LeRoy .
Double check this for me:
A player kicks a football with an initial velocity of 3.00 m/s at an angle of 60.0* above the horizontal. What is the horizontal distance traveled by the football?
s = vi*t + (1/2)(a)(t2)
0 = 3sin(30)t + 0.5(9.8)(t2)
t = 0.306 seconds
x = vt there is no horizontal acceleration
x = 3cos(30)(0.306 seconds)
x = 0.795011321 meters
x = 0.795 meters

That would not even get over the line of scrimmage for a PAT! You DID correctly solve the quadratic equation for t. You should have used sin 60 degrees in the s=0 equation and cos 60 degrees in the x equation. You got the right answer anyway, because sin (60) cos(60) appears in the answer,. and that equals sin(30) cos(30).
The horizontal distance travelled is
X = (V^2/g) sin 2A, where A = 60 degrees.
The problem with your low answer is that they (or you) probably made a typo error in the initial velocity of the kicked football. It is more likely to be 30 m/s.
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