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Note that there is no component of acceleration in the direction of the velocity--those two vectors are perpendicular. That remains true at all points in the path. Thus theall of the objects velocity is in the horizontal direction, and all of the acceleration is in the vertical direction.

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Chet

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Not true. For any arbitrarily small Δtime, the velocity in that direction changes. The force is instantaneously 0 in that direction but is nonzero at any Δtime before or after.all of the objects velocity is in the horizontal direction, and all of the acceleration is in the vertical direction..

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Realize that the OP was talking about one instant (the at the "top" of the orbit), not in general.Not true. For any arbitrarily small Δtime, the velocity in that direction changes. The force is instantaneously 0 in that direction but is nonzero at any Δtime before or after.

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The vertical acceleration is increasing the vertical speed (which is ~ 0.0) faster than the horizontal acceleration is decreasing the horizontal speed ( which is ~ total speed)I would think that the vertical acceleration is increasing the speed faster than the horizontal is decreasing the speed.

Yes. Sorry. I guess the real question in the OP is about the rate of change of the horizontal and vertical speeds. The answer to that is that the trade off is exactly so that the total speed remains constant.Realize that the OP was talking about one instant (the at the "top" of the orbit), not in general.

Speed

Because the component speeds are squared, the vertical speed, which is nearly zero, must increase faster than the horizontal speed, which is nearly the total speed, decreases.

Because the component forces vary proportional to sine and cosine, which trade off exactly right, the total speed remains constant.

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