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ilikebigbutts ilikebigbutts
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12 years ago
a person is standing on a table holding 2 identical balls. one ball he throws upward and at the same instants he drops  the second ball. will both balls will reach the ground with same final velocity. please give reason. an explanation or proof by numerical figures will be appreciated \thanks
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12 years ago
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#2
12 years ago
it depends upon the density and mass of the body
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#3
12 years ago
No. The acceleration on both balls will be the same because of gravity. Because the first ball is thrown upward, when it gets back to your hand it will be moving downward. When you drop the second ball, it starts from rest. Therefore, the first ball essentially has a head start in building up velocity due to it being thrown.
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#4
12 years ago
No. The ball that was dropped started at a given highth with a velocity of zero. The higher it is the more time it gets to accelerate before hitting the ground. The ball that was thrown up will slow down until it gets to a velocity of zero, by witch time it is much higher then the ball that was dropped. It has more time to accelerate downwards then the first ball because it was given extra highth.

Energy = meters from the ground times mass of the object. (E = m x kg)
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#5
12 years ago
No, they will not.  They start with the same potential energy, but with one ball you imparted energy to it by throwing it upward.  The dropped one will have initial total energy of mgh (mass * gravity * height), where as the total energy of the tossed one will have mgh + .

If you threw the still ball downward instead of dropping with the same amount of energy, then it would be the same velocity.

Another way to think of it is that the ball thrown upwards will not be starting at velocity of zero when it passes you on the way down, and it will still be accelerating.

It's just conservation of energy for each ball.

BTW, you asked about velocity, so it doesn't matter if the two exited the hands at the same time or not.

A good way to approach these type of questions is to take them to the extreme.  Imagine you dropped one ball a few inches off the floor and then tossed the other ball upwards from a few inches off the floor with a cannon so that it then careened to earth at maximum velocity.  Obviously the first would hit with a tap, and the other would hit significantly harder.
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#6
12 years ago
You got 2 good answers. Bottom line the ball thrown up will have a higher velocity than the dropped ball. Your statement is wrong. Hope your teacher did not tell you that or ask you to prove something that is incorrect
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#7
12 years ago
NO both the balls have different final velocities....
consider the third equation of motion  
v * v (square of final velocity) = u * u (square of initial velocity) + 2gh
h: height covered while falling

here in this case
-the ball which is dropped has u = 0 and is at the height suppose "H"
-for the second ball , it first goes upwards to a maximum height let "x" and then comes down to the ground which is now at the height of (H +x) from that point.

now considering both the points having initial velocity 0 , then the third equation can be written as
 v * v (square of final velocity) =  2gh
therefore ,
v * v (square of final velocity) is directly proportional to "h"   that is the height covered.

since the height ( H + x ) is clearly greater than the H  therefore the final velocity for the ball which one is thrown upwards will be more.

HOPE YOU UNDERSTAND WHAT I WANT TO SAY . SORRY FOR ANY MISTAKE
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#8
12 years ago
Say, height of table =h   height from where the other ball started downward motion =H

       final velocity of the ball dropped =v1   final velocityof ball thrown upward= v2

v1^2 = o + 2gh =2gh  and  v2^2 =0+2gH =2gH

As  h > H   hence,    v1>v2
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