Velocity of Thrown Ball: 2.0m/s - Physics Q&A

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When a ball is thrown upward at a velocity of 2.0 m/s, it momentarily reaches a velocity of 0 m/s at the peak of its trajectory before descending. Upon returning to the original height, its final velocity will be -2.0 m/s, indicating the same speed but in the opposite direction. The ball experiences a constant acceleration of -9.80 m/s² due to gravity throughout its motion. This means that while the magnitudes of the initial and final velocities are equal, their signs differ because velocity is a vector quantity. Thus, the ball does indeed come down at the same rate it was thrown up, assuming no external forces act on it.
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If I was to throw a ball up in the air at a velocity of 2.0m/s, would it come down at the same rate?

For example, after the ball pauses for that split second and then turns downward, would it be going at -2.0m/s?

I know it would be accelerating at -9.80 m/s(squared) , but would it's initial or final velocity be equal to -2.0m/s?

Thanks!
 
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ThomasMagnus said:
If I was to throw a ball up in the air at a velocity of 2.0m/s, would it come down at the same rate?

For example, after the ball pauses for that split second and then turns downward, would it be going at -2.0m/s?

I know it would be accelerating at -9.80 m/s(squared) , but would its initial or final velocity be equal to -2.0m/s?

Thanks!

At the split second the ball reaches the top and then starts to fall again its velocity would be zero. It would continue to accelerate back to its initial velocity at the point of release.

So if its initial release velocity was 2 m/s it would travel up until it stopped and changed direction and return to your hand with a final velocity of -2 m/s (note the sign change due to velocity being a vector, but the magnitude is the same - neglecting any losses).

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