Free Fall Acceleration Problem

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The discussion revolves around a physics problem involving a scientific balloon and a free-falling instrument package. The maximum height above the break-free point is determined to be 19.6 m. For part B, the user struggles to find the correct height of the break-free point above the ground, using the kinematic equation but arriving at an incorrect answer. The time at which the package reaches its maximum height is identified as t = 4 s, where its velocity is momentarily zero. The conversation emphasizes the importance of analyzing the velocity-time graph to solve the problem accurately.
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Homework Statement



As a runaway scientific balloon ascends at 19.6 m/s, one of its instrument packages breaks free of a harness and free-falls. The figure below gives the vertical velocity of the package versus time, from before it breaks free to when it reaches the ground.
2-p-053.gif


(a) What maximum height above the break-free point does it rise?
(b) How high is the break-free point above the ground?


Homework Equations





The Attempt at a Solution



i know part A is 19.6 m
i just can't seem to get the correct answer for part B

these are the variables that i have and i think i'll need
V0=0 m/s
t=6.0 s
a=-9.8 m/s/s
y-y0=?

i used this kinematic equation
y-y0=vt-1/2at^2
but the answer still isn't correct
 
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The package is at the break-free point at time t = 2 s and it is moving up. Can you figure out by looking at the graph at what later time it is at the same point, moving down?
 
it's at the same point at t=4 s
 
At t = 4 s the velocity is instantaneously zero. This mean maximum height.
 
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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