One-Dimensional Kinematics: Height vs Time

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The discussion revolves around solving a one-dimensional kinematics problem involving height and time on the fictional planet Zoot. Participants clarify that the standard free-fall acceleration of 9.8 m/s^2 does not apply in this scenario, prompting the need to determine the local acceleration due to gravity. To find this, the equation s = 1/2 at^2 + u*t can be used, particularly focusing on the second half of the graph where initial velocity is zero. Additionally, the final velocity can be calculated using u^2 = v^2 - 2*a*s, with the first half of the graph providing the necessary values. The exchange highlights the importance of context in physics problems and the collaborative nature of problem-solving.
rdn98
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This problem has a picture. You can view it at http://www.geocities.com/rockdog_84/One.htm

Ok. Maybe I'm missing something here, but for part one, I thought free-fall acceleration would be 9.8 m/s^2. Of course, computer didn't take it, so is there a way to find it from the graph?

As for the second part of the question, I thought initial velocity is 0, but its not the right answer. :-(
I guess another way to find initial velocity would be take the derivative of x(t) when t = 0, but I don't know the equation for this problem.

Any insight is appreciated.
 
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9.8 m/s^2 is the correct acceleration due to gravity on Earth, however the scenario in question takes place on the planet Zoot. If a question does not take place on Earth, then that is a big red flag that acceleration due to gravity is NOT 9.8 m/s^2.

You can figure out g by using s = 1/2 at^2 + u*t (Take the second half of the graph where u is 0).

Once you know g, you can use u^2 = v^2 - 2*a*s to find the final velocity. (Take the first half of the graph this time so v = 0)
Note: -
u = initial velocity
v = final velocity
t = time
a = acceleration
s = displacement

Claude.
 
big thanks!

Man, I would never have thought of that! Thank you so much.
 
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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