Calculating Acceleration of a Sphere Thrown Upward

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The discussion revolves around calculating the acceleration of a sphere thrown upward from a height of 13 meters with an initial velocity of 20 m/s. Participants clarify that the acceleration due to gravity is approximately 9.8 m/s², which is the correct answer, rather than the initial velocity. The height of 13 meters does not affect the magnitude of acceleration, which remains constant during the motion. There is confusion between acceleration and velocity, with one participant mistakenly equating the two. Ultimately, the consensus is that the magnitude of acceleration after 1.5 seconds is 9.8 m/s².
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I know I already have a topic for a previous question, but here's another simple one, but I think I am just missing some simple equations.

"What is the magnitude of acceleration after 1.5 seconds of a sphere thrown upward from a height of 13 meters at 20 m/s^2?"

Choices
a) 5.3 m/s^2
b) 9.8 m/s^2
c) 15 m/s^2
d) 32 m/s^2
e) 390 m/s^2

I realize these are quite obvious answer choices, and the answer is most likely D, but I don't know how to get it. I only have the "four kinematics" equations to work with, but I think there are other equations I can use for this. How does the height of 13m effect my answer, if it indeed does.
 
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Use the equation y = y_{0} + v_{y0}t + \frac{1}{2}a_{y}t^{2}
 
courtrigrad said:
Use the equation y = y_{0} + v_{y0}t + \frac{1}{2}a_{y}t^{2}

ok, using that equation, doesn't it leave me with two unknowns? I have time, initial distance in y, and initial velocity in y. I don't have the total distance in y, or the acceleration, which I am trying to find. Is that right, or am I mistaken...I might need to find total distance first, and then the acceleration.
 
ch3570r said:
I know I already have a topic for a previous question, but here's another simple one, but I think I am just missing some simple equations.

"What is the magnitude of acceleration after 1.5 seconds of a sphere thrown upward from a height of 13 meters at 20 m/s^2?"

Choices
a) 5.3 m/s^2
b) 9.8 m/s^2
c) 15 m/s^2
d) 32 m/s^2
e) 390 m/s^2

I realize these are quite obvious answer choices, and the answer is most likely D, but I don't know how to get it. I only have the "four kinematics" equations to work with, but I think there are other equations I can use for this. How does the height of 13m effect my answer, if it indeed does.
What causes the object to accelerate? What is the magnitude of this acceleration? This is more of a "did you memorize this constant?" kind of question than a problem about applying kinematics equations. Lastly why do you think it's D?
 
well, the acceleration down would be gravity (9.8 m/s^2)...wait, so the answer is B!? I was thinking that because the object is traveling at 20m/s^2, after 1.5 seconds, it would be slighty over 20m, and the closest answer is D. But I guess if its asking for the "magnitude" of the acceleration, it would be gravity.
 
ch3570r said:
well, the acceleration down would be gravity (9.8 m/s^2)...wait, so the answer is B!? I was thinking that because the object is traveling at 20m/s^2, after 1.5 seconds, it would be slighty over 20m, and the closest answer is D. But I guess if its asking for the "magnitude" of the acceleration, it would be gravity.

Yes the answer is B. And no after 1.5 seconds the object is not traveling at 20m/s^2 that is an acceleration not a velocity! Acceleration is measure in units of meteres per second squared and near the surface of the Earth the value of acceleration due to gravity is nearly constant at 9.8m/s^2.
 
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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