Projectile: Do not understand wording

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SUMMARY

The discussion revolves around a physics problem involving projectile motion, specifically how far in advance a runner should release a water cup to reach a garbage pail located 0.50 m away horizontally. The runner's speed is given as 6.20 m/s, and the solution indicates that the correct answer is 2.0 m. Participants express confusion regarding the phrasing "how far in advance," clarifying that it refers to the horizontal distance needed for the cup to reach the pail, rather than the vertical distance.

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Homework Statement



Participants in a road race take water from a refreshment station and throw their empty cups away further down the course. If a runner has a forward speed of 6.20 m/s, how far in advance of a garbage pail should he release his water cup if the horizontal distance between the lid of the garbage can and the runner\'s point of release is 0.50 m?

Homework Equations



Rx = Rcos(theta)
Ry = Rsin(thetha)
Kinematics equations

The Attempt at a Solution



I don\'t know what exactly this question is asking for. First of all, what does it mean by \"how far in advance\"? Normally, I would assume that this would mean the horizontal component of distance. However, this is certainly not the case as horizontal distance upon the point of release is given in the question. Unless if this cup is going to magically travel beside him in the x direction after release, it makes no sense whatsoever, especially considering he would merely drop the cup on the ground in reality, 0.50 m short of the garbage can.

By the way, the answer in the back is 2.0 m.
 
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i'm not sure, but it sounds like the problem means the vertical distance is .5m
 

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