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Old Aug 5th 2010, 08:08 AM   #1
ohm
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graph of motion under gravity

consider a ball is released from a height h, it strikes the ground and rebounces by loosing 10% of its velocity at each rebounces. I have to plot motion of this ball between t1 to t2 sec.
At first strike , it has two vel one with which it strike(v=sq root of 2gh) the ground and one with which it rebounces(i.e. 10%less that v). Suppose at time t ball strikes, then which velocity should I take at time t(vel at which each rebounces or velocity at which it strikes) and why?
Answer is given vel at which it rebounces.
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Old Aug 5th 2010, 01:02 PM   #2
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At time t=T1 (when the ball hits the ground) it will have a downward (negative) velocity equal to sqrt(2gh), as you said, the instant prior to hitting the ground. Your graph of the ball's position will have a line sloping downward right up to to t = T1. Then it rebounds, with an initial velocity upward of 90% of sqrt(2gh). So the velocity curve is discontinuous at t=T1. If you plot velocity versus time the curve has a big discontinuity at the instant the ball strikes the ground, shifting form v = -sqrt(2gh) to +0.9sqrt(2gh) instantaneously.
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Old Aug 6th 2010, 01:07 AM   #3
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you mean right at the time t=T1, plot will have two points ,one at v=-sqrt(2gh) other at v=+0.9sqrt(2gh). Can a body have two velocity at the same time?
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Old Aug 6th 2010, 05:19 AM   #4
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Originally Posted by ohm View Post
you mean right at the time t=T1, plot will have two points ,one at v=-sqrt(2gh) other at v=+0.9sqrt(2gh). Can a body have two velocity at the same time?
Thanks
No. There is a fraction of a second time delay between re-bouncing and striking.
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Old Aug 6th 2010, 06:18 AM   #5
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Take a look at the attached charts, showing the position and velocity of a bouncing ball. Note that at the instant the ball hits the floor it has a big instantaneous change of velocity. This is an idealized graph - such a change in acceleration implies an infinite acceleration at that instant. If you consider what happens to a real ball, it compresses as it hits, taking some amount of time to fully compress. At that instant its velocity is zero. During this compression the ball is deceleraing very rapidly, but its not infinite. Then it starts to spring back, which causes it to accelerate upward.


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Old Aug 6th 2010, 08:30 AM   #6
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sorry! Where is attached chart?
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Old Aug 6th 2010, 08:57 AM   #7
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Originally Posted by ohm View Post
sorry! Where is attached chart?
Hmm.. you don't see 2 charts embedded in my previous post? I do...

Try this:

http://i885.photobucket.com/albums/a...ncing_Ball.jpg


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Old Aug 6th 2010, 09:38 AM   #8
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sorry! Due to some technical problem in my internet connection, I was unable to see the plot.
I think the graph represents the motion of a body projected verticaly upward with an initial velocity 9.8m/s comes at rest after 1sec. At t= 2sec it has two velocity -10 m/s and 9m/s. Can a body have two velocity at same time ?
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Old Aug 6th 2010, 11:00 AM   #9
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A real object can not have two velocities at the same time. But then a real object wouldn't instantaneously bounce with infinite aceleration the way the curve shows (recall that infinite acceleration requires infinite force, so it can't happen). So this is more of a math question than a physics question. Basically for the velocity curve that I drew the exact value at t = 2 seconds is undefined. Mathemeticians say that the limit of the curve v as t approaches 2 seconds from the left is -9.8 m/s, and the limit of the curve v as t approaches 2 seconds from the right is +8.82 m/s. Which means depending on how you look at it, the velocity at t= 2 seconds is either value, hence it's undefined. You can only talk about the value of v just slightly before time t = 2, or just slightly after.
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Old Aug 7th 2010, 01:38 AM   #10
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what has made acc infinite here? Since change in vel is finite here, so is it time that has made it infinite. If yes ! Then can't I measur time here. Aslo you have wrote that when ball compress it deaccelerates but it is not infinite here. So what is the factor here that make this difference?
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