Projectile: Do not understand wording

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The discussion centers on a physics problem involving a runner throwing a cup towards a garbage pail while moving at a speed of 6.20 m/s. Participants express confusion over the phrase "how far in advance," questioning whether it refers to horizontal distance, as the problem already specifies a horizontal distance of 0.50 m. There is skepticism about the cup's trajectory, with concerns that it would not travel horizontally alongside the runner after being released. One participant suggests that the answer provided in the textbook, 2.0 m, may imply a vertical distance instead. Overall, clarity on the problem's wording and intended meaning is needed for accurate problem-solving.
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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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