What is the final speed of a ball thrown from a window 3.6m above the ground?

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SUMMARY

The final speed of a ball thrown from a window 3.6 meters above the ground with an initial speed of 2.8 m/s is calculated using the kinematic equation v² = v₀² + 2ad. The correct final speed upon impact is 8.85 m/s, assuming a downward acceleration of 9.8 m/s². The height from which the ball is thrown does not change the final speed, as the vertical motion is independent of the initial upward velocity.

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Kinematics Question please help.

Homework Statement


a ball is thrown vertically upward from a window that is 3.6m above the ground. Its initial speed is 2.8m/s . What speed will the ball hit the ground.
show what formulas were used please.

Homework Equations


v2^2=v1^2+2ad

The Attempt at a Solution



im wondering, wouldn't the height be different, because it is thrown vertically up?
so the distance would be different, changing the speed (Vf)?

I get 8.85m/s when i assume d=3.6m Vo=2.8m/s a=9.8m/s
wouldnt that be the speed of the object if it were thrown to the ground?
 
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anything? can anyone even just say whether or not my answer is right?
 

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