Overlooking something, Need Help

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The discussion revolves around two physics problems involving an Olympic diver and a baseball. For the diver, the key points include calculating the final velocity upon hitting the water and determining the average acceleration while submerged. The baseball problem involves finding the velocities at the top and bottom of a window and the height from which it was dropped. The participant initially struggled with these calculations but received guidance on the correct approach. Ultimately, they were able to identify their oversight and make progress in solving the problems.
timpn86
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Homework Statement


2. An Olympic diver steps off of a 10. m high diving board, plunges into the water and comes to rest beneath the surface of the water 1.7 s after entering it. a) Determine the velocity at which the diver hits the water. b) Determine the average acceleration of the diver in the water.

Homework Equations



i got to determine the final velocity at which the swimmer reaches the water that in turn will be the initial velocity at which the swimmer enters the water.

equations are attached

The Attempt at a Solution



Falling
vi = 0 m/s
vf =
a = 9.8 m/s2
t =
s = 10m

In water

Vi =
Vf = 0 m/s
a =
t = 1.7 s
s =

Homework Statement



A baseball is observed to drop downwards past a 3.0 ft long open window in 0.20 s. a) Find its velocity at the top of the window. b) Find its velocity at the bottom of the window. c) Find how high above the top of the window the object was dropped.

Homework Equations



equations attached

The Attempt at a Solution



3ft x .3048 = .9144m

From Drop
vi = 0m/s
vf =
a = 9.8m/s2
t =
s =



Top of Window
vi =
vf =
a = 9.8m/s2
t = .2s
s = .9144m


i was able to finish all 7 other problems on the page but these 2. What am i overlooking?
 

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Hi timpm,welcome to PF.
Since you have solved the rest of the problem, show some attempt of these two problems. Let me see wher you are stuck.
 
for first problem, i think you are proceeding in right direction. since height 10m is given, you can find the final velocity before striking water. let it be V.

so avg. acceleration is V/1.7
 
haha thanks rl.bhat... upon reviewing the problem thurther i was able to find what i missed.
 
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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