How Fast Was the Steel Ball Rolling?

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The problem involves a steel ball rolling off a table 1.063 m high and landing 0.229 m horizontally from the edge. The time of fall was calculated to be approximately 0.4657 seconds using the equation for free fall. The horizontal velocity was then derived using the formula v_x = Δx/t, leading to a final velocity of about 0.4916 m/s. Initially, there was confusion over the calculations, but it was later confirmed that the initial answer was indeed correct. The discussion highlights the importance of careful calculation and verification in physics problems.
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Homework Statement



A steel ball rolls with a constant velocity across a table top 1.063 m high. It rolls off and hits the ground +0.229 m horizontally from the edge of the table. How fast was the ball rolling?


Homework Equations



d=.5at^2
X=Vxt

The Attempt at a Solution



I first did
1.063=.5(9.8)t^2
got the time to be .4657 seconds
then put it in the X=Vxt
and got Vx to be .4916, and I am not sure where to go from there
 
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You approached the question correctly, but your final result is wrong.

Using \Delta x=v_{x}t where:
\Delta x=displacement
v_{x}=horizontal velocity
t=time

the displacement will be the distance it traveled from the edge of the table, the time is what you just calculated, and all that is left is the horizontal velocity.

Rearrange the formula to make velocity the subject: v_{x}=\frac{\Delta x}{t} and substitute to find your answer.

EDIT: Sorry your answer was correct, calculator error on my part.
That is the answer to the question, in metres/second (ms-1)
 
Last edited:
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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