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Velocity of throwing a ball

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  • #1
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Homework Statement



You toss a ball up in the air with an initial velocity of 6 m/s (immediately after leaving your hand). Use gravity = -9.8 m/s2.

What is its acceleration immediately after releasing the ball?
What is its velocity at its peak height?
What is its acceleration at its peak height?
What is the ball's velocity when it falls past the point at which you released it?

Homework Equations



d=v0t+1/2gt^2
v=v0+at
v^2-v0^2=2gd

The Attempt at a Solution



I got the first three answers (-9.8, 0, -9.8, respectively), but I'm confused about the last question. I tried solving for distance of the ball going up to the top and tried to use that....I also tried plugging in the numbers for the equations with v0=0m/s b/c that was the velocity at the top of the arc...I'm not sure if I'm even on the right track at this point...
 

Answers and Replies

  • #2
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- The first three answers are correct except for missing the units (which is an error that in an exam I'd give you minus points for).
- Do you understand the 3rd equation you stated? I don't, but you can use it to almost directly get the missing answer you are looking for (and get the correct result). If you don't understand the equation yourself, then either try understading it or simply use the first two equations and just try to find as much of the values a,d,v,t for the three relevant points (starting point, maximum height and point of return) - you can ultimately get all of them. This way you also get the answer you are looking for, you just might end up things that were not asked for as intermediate steps which is not really a problem and possibly even a good exercise.
 
  • #3
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sorry about the units (I didn't have to write them...they were already preprinted)

I'm still kind of confused about solving the problem. I tried breaking it up into portions like you suggested, but the most I figured out was that when the ball makes a complete arc, its velocity, right as it reaches your hand should be equal to the beginning velocity. When they ask for the velocity past that point, I'm getting a little bit confused. How can I solve for that velocity if I don't know the distance from the beginning point to the end point? I know that I can solve for time, but I don't know how long it takes for the ball to go past that point and I can solve for it, but I would need to know the distance of how far the ball went from that horizontal start point. Am I just overthinking this problem? (or am I just completely way off base?)
 
  • #4
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35
I'm not a native english speaker but I understand "when it falls past the point at which you released it" as "when it's at the height of the hand again". The velocity is, similarly as you mentioned, the negative of the inital velocity, then (it's fyling in the opposite direction, of course). You could get the velocity for any distance below your hand by determining the time it takes to fall down that distance with eq (1) and then get the velocity at this point with eq (2). Keeping the distance as a variable, you can even get a generic expression v(d). But I from the level that the other questions are, I don't think that's what is asked for - I think the -6 m/s is what is asked for.
 
  • #5
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oh, okay....that makes a lot of sense...I guess that I was just overthinking the problem...thank you so much for your help!
 

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