Does the rock travel 321m vertically when kicked off a bridge?

AI Thread Summary
The rock does travel 321 meters vertically in free fall, regardless of whether it is kicked horizontally or dropped. The vertical distance remains constant at 321 meters, as it must reach the river's surface. The trajectory of the rock will follow a parabolic path due to its horizontal velocity, but this does not change the vertical distance it falls. The horizontal motion does not affect the vertical distance; thus, the rock's vertical displacement is indeed 321 meters. Understanding this concept is crucial for accurately analyzing projectile motion.
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A bridge rises 321m above the Arkansas river. Suppose you kick a rock horizontally off the bridge. The magnitude of the rock's horizontal displacement is 45.0 m. Fine the speed at which the rock was kicked. (This is not what i ask you here)

Does the ball travel 321m vertically in free fall? If so, then i have a question, we know that 321m is a vertical path straight down to the river, which is perpendicular. but when we draw the path of the ball that travels is half of a parabola, which is a curve, so if we have the common sense we know that if we measure the distances of these two paths, they are different. THe curve is longer than the straight line. so i don't know if it travels 321m vertically in free fall.
Hope you can explain it to me, thank you.
 
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I don't really understand your question, but I'll try to explain a little.
It does travel 321 m in free fall. It cannot travel less (or it would hang above the river) or more (it would free-fall inside the river). It doesn't matter if you kick it off horiontally or drop it over the edge. Assuming no friction, the horizontal velocity of the rock will not change. That is, if you drop it, it will fall straight down and plunge into the water with a velocity of 9.81 t_\mathrm{fall} aimed straight down; if you kick it off at 20 m/s it will come down with a velocity that is aimed slightly downwards, slightly forwards (in fact, it will have a horizontal component of 20 m/s, and a vertical component of 9.81 t_\mathrm{fall}). You should add them vectorially (if you know how to do that), or you can draw it in a picture by drawing the horizontal and vertical lines (on scale) - then the final velocity will be the tip of the parallelogram spanned by those.
 
ok, i mean that the pictures is like this:
1)if there is no horizontal velocity, it would go straight down perpendicularly:
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2) However, there is horizonal velocity involved in this question, so it travels down like half of a parabola, a curve
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/ (sorry, i can't show it having a slope, so just imagine it. then, this is a curve, does the ball still travel 321m straight down?) if you still don't understand what i mean, here: a straight line between two points is the shortest if we measure, a curve between the same two points would be longer than the straight line if you use ruler to measure it mathematically.
 
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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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