How Do Final Speeds of Two Balls Compare When Thrown from a Building?

AI Thread Summary
When a student throws ball 1 upward and ball 2 downward from a building, both are released with the same initial speed, vi. Ball 1 will ascend before descending, while ball 2 falls directly downward. As ball 1 reaches the thrower's height again, it will have the same speed, vi, but in the opposite direction. Both balls experience the same gravitational acceleration, leading to the conclusion that their final speeds upon reaching the ground will be equal. Therefore, v1f equals v2f, confirming that both balls hit the ground with the same final speed.
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Homework Statement


A student at the top of a building of height h throws ball 1 upward with a speed of vi and then throws ball 2 downward with the same initial speed, vi. How does the final speed of ball 1, v1f, compare to the final speed of ball 2, v2f, when they reach the ground?
v1f = ______ multiplied by v2f

Homework Equations


a = delta v/delta t
vf = vi + a(delta t)

The Attempt at a Solution


I drew a picture showing the person throwing ball 1 upward and then it descending and ball two falling to the ground.
 
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Okay well let's think about this. We know if there is presence of uniform force in only one dimension then the trajectory is going to be parabolic. For ball 1, it will leave his hand at vi go up in its arc, reach a maximum height (where vertical velocity is zero) begin to fall again with identical acceleration as before.
This implies that when it passes the thrower for, it will have the same velocity it was thrown with, but in the opposite direction.
 

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