physics repost
posted by Jon .
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?
2(3.00)(cos60)/9.80
(3.00)(cos60)(.530219635)
0.795m
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"The looks like a pretty short punt to me: about 2 feet. The correct formula is (horizontal velocity component)* (time of flight). There should be a sin 60 and a v^2 in the answer, besides the factors already there in your first equation."
Ok but I that's the biggest answer choice I have.
A)0.312m
B)0.397m
C)0.673
D)0.795 what I got
I don't understand "The correct formula is (horizontal velocity component)* (time of flight). There should be a sin 60 and a v^2 in the answer, besides the factors already there in your first equation."

If you don't understand the statement, then you don't understand the physics.
The physics is this: The ball is in the air for some time. The question is how long?
finalheight= initialheight+ initial velocity*timeinair 1/2 g timinair^2
The final and initial height is zero. You need the time of flight.
0=0+3sin60*t 4.8t^2
t=0 or t=sqrt (3sin60/4.8) check that.
horizontal distance= 3cos60*time, which I don't get your answer.
I agree with your answer.