Calculating Height & Time of a Falling Ball | Physics Problem Solution

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The discussion focuses on solving a physics problem involving a ball dropped from rest, covering three-quarters of the distance to the ground in the last second of its fall. Participants emphasize identifying the type of motion and relevant equations, specifically one-dimensional motion in the vertical direction. Key equations mentioned include those for velocity and position, with a focus on incorporating the "last second" aspect to determine the height and total time of the fall. The approach involves calculating the time to fall from the initial height and adjusting for the last second of the fall. This method leads to a solution for both the height from which the ball was dropped and the total time taken for the fall.
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A ball, dropped from rest, covers three-quarters of the distance to the ground in the last second of its fall. a) From what height was the ball dropped? b) What was the total time of the fall?

Hey, I don't know how to approach these types of problems. Help?
 
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Hi Manni. The approach is nearly always to ask, "What type of motion is it? What formulas apply?" Once you can write a formula for the height as a function of time, you'll be able to apply a bit of math to it and find the answers!
 
It's 1 dimensional motion in the y-direction. As for the equations, there's 2 that I can think of off the top of my head:

Velocity = v(initial) + at
Position = x(initial) + (0.5)[ v(initial) + v ]t

I don't know what to do from there.
 
Position = x(initial) + (0.5)[ v(initial) + v ]t
should be y = yi + Vi*t + ½ a⋅t²

How do we put in
A ball, dropped from rest, covers three-quarters of the distance to the ground in the last second of its fall.
Have to do something with "last second" for starters!
Maybe find the time to fall from height Yi. Then subtract 1 to get the time at the point where it still has 3/4 of the distance to fall.

To find the time to fall, work with the second equation and solve for t.
 
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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