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

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SUMMARY

When a ball is thrown upward at a velocity of 2.0 m/s, it reaches a peak height where its velocity becomes 0 m/s before descending. Upon returning to the original release point, the ball's final velocity will be -2.0 m/s, indicating a change in direction but maintaining the same magnitude of velocity. The acceleration due to gravity is consistently -9.80 m/s² throughout the motion, affecting the ball's velocity as it ascends and descends.

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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).

CS
 

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