Solve Soccer Ball Kick Homework Problem: 13.68m?

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The homework problem involves calculating the distance from which a soccer ball is kicked to reach a height of 11.5 meters at its peak. Initial calculations yielded a distance of 13.68 meters, but discrepancies arose due to variations in time calculations, with some suggesting 1.53 seconds instead of 1.52. The vertical component of the kick was also discussed, revealing that using gravity as 10 m/s² led to a distance of 14.04 meters, which aligns closely with the book's answer of 14.0 meters. The discussion highlights inconsistencies in the problem's parameters, indicating potential errors in the original question. Clarification will be sought from the teacher regarding these inconsistencies.
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Homework Statement


Your friend is standing on the roof of his house, which is 11.5 meters off the ground. You are standing some distance away from the house and you kick a soccer ball at a total initial velocity of 18.0 m/s @ 60 degrees above the horizontal from ground level. If the soccer ball gets to the top of the house at the peak of its motion, how far away were you standing?


Homework Equations


d=1/2g*t^2
d=vt

The Attempt at a Solution


I already posted this once and got my mistake explained to me, so I tried again. First I found the time by doing d=1/2g*t^2 and got the time to be apporximately 1.52 seconds. Then I found the horizontal component, which would be 9 since the resultant and components form a 30-60-90 triangle. I multiplied 9 * 1.52 to get 13.68 meters, but my book says that this is wrong. Could anyone explain what I'm doing wrong and what I should change?
 
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Solving the problem your way, t is actually 1.53 s, not 1.52, which yields 13.79 m instead of 13.68. Is that the problem?

If it's not, the 11.5 m is actually redundant data, because the question also tells you the ball reaches your friend at the peak of its motion. You can calculate time from that easily enough: just divide the vertical component of the ball's speed by g. You get 1.79 s, which isn't consistent with the height given by the question. (It should be; the question is simply inconsistent.) If you use that to calculate distance, you get a different answer. Is this the problem?
 
I tried the second way you described (Vertical Component/Gravity=Time) and got 1.56 (forgot to mention that we're supposed to use 10m/s^2 as gravity due to acceleration rather than 9.8) Multiplied by 9 that's 14.04, and the book says the answer is 14.0 so I believe that that's the answer. Thanks! I'll mention that the problem is inconsistent to my teacher. The problems were made by him so he might have messed up. Thanks again!
 
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Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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