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

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The discussion revolves around calculating the final speed of a ball thrown vertically from a height of 3.6 meters with an initial speed of 2.8 m/s. The relevant kinematic equation used is v2^2 = v1^2 + 2ad, where 'a' is the acceleration due to gravity at 9.8 m/s². There is confusion regarding whether the height should be adjusted since the ball is thrown upward before descending. The calculated final speed is 8.85 m/s, assuming the distance is 3.6 m, but there is uncertainty about whether this speed is accurate given the initial upward motion. The thread seeks confirmation on the correctness of this calculation.
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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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