Free Fall Object: Time and Height Calculations from Rest | Homework Equations

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To solve the problem of an object in free fall that travels half its total distance in the last second, it is essential to use the equation for constant acceleration, d = V1Δt + 1/2 at². The acceleration due to gravity is -9.8 m/s², and the initial velocity (Vi) is 0 m/s. The approach involves recognizing that the time taken for the first half of the fall equals the total time minus one second. Calculating the distances for various times revealed inconsistencies, suggesting the need for a velocity vs. time graph to determine displacement accurately. This method allows for a clearer understanding of the relationship between time, height, and distance in free fall scenarios.
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Homework Statement



If an object travels half its total path in the last second of its free fall, from rest, find:
(a) the time, and
(b) the height of the fall


Homework Equations



I'm not sure at all what equations to use

The Attempt at a Solution



The only thing I can think of is just assuming a mass of an object, and guessing the time, but I didn't have any luck with this
 
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an object in free fall is just a one-dimensional constant acceleration problem with an acceleration of -9.8 m/s2

If you recall your equations of one-dimensional constant acceleration, you should remember that mass does not play a role. At least in a simplified problem like this one.

If the object falls half its total distance traveled in the last second, that means that it also fell the first half of its distance in a time equal to the total time minus one second
 
Would a good equation for this be:
d=V1Δt+ 1/2 at2
 
steven1495 said:
Would a good equation for this be:
d=V1Δt+ 1/2 at2

Yes, that's equation for constant acceleration.
For easier calculation, assign acceleration as positive.

Write down all the known data and the unknown data.
If there are 3 unknown data, then you have to find 3 independent equations.
 
the known data is
a= -9.800 m/s^2
Vi= 0 m/s

I used the above equation to find the distances with the times, but all the times I did (up to 6 seconds) didn't work, because after 6 seconds, the distance for the last second was less than the sum of all the other distances
 
Maybe you sketch a velocity vs. time graph.
The area under the graph is the displacement.
 
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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