One-Dimensional Motion with Constant Acceleration

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The discussion revolves around calculating the distance a rock falls in the first 7 seconds of free fall from a cliff, using the equation X = Xo + VoT + 1/2AT^2. The initial position (Xo) can be set to zero at the top of the cliff, simplifying the calculation since the difference in position is what's important. The initial velocity (Vo) is zero because the rock is released from rest. The acceleration due to gravity (g) can be treated as either positive or negative, depending on the chosen coordinate system. Understanding the underlying principles of motion is emphasized over rote memorization of formulas.
PhysicallyAbel
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Homework Statement


A rock is released from rest from the top of a very high cliff, and accelerates downward at g.
Approximately how far does the rock travel in the first 7 seconds of its free-fall? (Assume no air friction.)

Homework Equations


X=Xo+VoT+1/2AT^2

The Attempt at a Solution


I understand how to work out the problem, but I do not know why it works. You simply substitute the numbers into get the distance. But for some reason, you are not given (nor do you use) initial position like the equation calls for. Can someone explain why it's simply thrown away? The initial position can't be 0, as it's being dropped from a mountain.[/B]
 
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PhysicallyAbel said:

Homework Statement


A rock is released from rest from the top of a very high cliff, and accelerates downward at g.
Approximately how far does the rock travel in the first 7 seconds of its free-fall? (Assume no air friction.)

Homework Equations


X=Xo+VoT+1/2AT^2

The Attempt at a Solution


I understand how to work out the problem, but I do not know why it works. You simply substitute the numbers into get the distance. But for some reason, you are not given (nor do you use) initial position like the equation calls for. Can someone explain why it's simply thrown away? The initial position can't be 0, as it's being dropped from a mountain.[/B]

The question is asking how far the rock travels. That's the difference between the final position and the initial position. ##x_0## cancels when you take the difference.
 
It sounds like you are memorizing formulas without really understanding them. That's a bad idea. You are free to take the "0" point any where you want. I suggest taking x= 0 at the top of the cliff. Further, the fact that rock is "released from rest" tells you that the initial velocity is 0. Finally, you can take the acceleration due to gravity to be either negative or positive- as long as you interpret the results in the same way- you are free to choose "positive" or "negative" either up or down.
 
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You can ignore the mountain altogether : you have a ( constant ) acceleration rate, a starting velocity (u = 0) and a time ( 7 seconds ), so use Newtons rules of motion.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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