Posted by **Jon** on Thursday, April 3, 2008 at 9:21am.

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

___________________________________

"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 thats 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."

- physics repost -
**bobpursley**, Thursday, April 3, 2008 at 10:03am
If you dont understand the statement, then you dont 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 dont get your answer.

I agree with your answer.

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