Objecs falling from different heights question

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When two objects, A at 100 meters and B at 1 meter, are released simultaneously, A will hit the ground first due to its higher velocity gained during its fall. The calculations confirm that A takes approximately 4.518 seconds to fall 100 meters, while B, starting from rest, takes about 0.452 seconds for its 1 meter drop. Even considering the effects of gravity, both objects experience the same acceleration, but A's initial velocity gives it a significant advantage. The only scenario where B could reach the ground first would require it to start with a higher velocity than A, which is not the case here. Therefore, A will always reach the ground before B under the given conditions.
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Let's say we have 2 objects. "A" is 100 meters above the ground. "B" is one meter above the ground. Release object A. When A reaches B, B is released and falls too.

Question... Do they hit the ground at the same time? Or does A get there first? Air resistance is not an issue... just the effects of gravity.

Thanks.
 
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I would summise (maybe incorrectly) that A gets there first as it has already built up significant velocity by the time it has traveled 99m. where as B has at the same point only started to accelerate.

Dave
 
liometopum said:
Let's say we have 2 objects. "A" is 100 meters above the ground. "B" is one meter above the ground. Release object A. When A reaches B, B is released and falls too.

Question... Do they hit the ground at the same time? Or does A get there first? Air resistance is not an issue... just the effects of gravity.

Thanks.

A reaches there first as rightly said by davenn since a would already have gathered a lot of velocity by the time it reaches to the height at which B is.
 
Does anyone have the numbers to tell for certain which hits first: A or B?
 
Assuming the objects are identical, there's no need to 'run numbers'.

There are three possibilities:

1. The objects are fairly aerodynamic with a terminal velocity which requires more than 100m under g acceleration to reach - in which case, A will reach the ground first as it has a significantly higher velocity than B over the final 1m (even if B didn't have to accelerate and just started out at it's maximum possible velocity it could achieve after 1m of acceleration it would still be traveling slower than A because A has had more than 1m over which to accelerate at 1g).

2. The objects are not so aerodynamic and terminal velocity is reached at any point between 1m and 100m - as above, A will reach the ground first as it has a higher velocity over the final 1m (again, even if B didn't have to accelerate and just started out at it's maximum possible velocity it could achieve after 1m of acceleration it would still be traveling slower than A because A has had more than 1m over which to accelerate at 1g).

3. The objects are not aerodynamic and reach terminal velocity within the first 1m of the drop - this means the final 99m are irrelevant to A's descent as they are at constant velocity. However, A will still reach the ground first as it's velocity over the final 1m will be constant whereas B will be accelerating for a period of it giving a lower average velocity over that 1m than A, which leads to a higher transit time for that distance.

The only way B could reach the ground before A would be for it to instantly be at a higher velocity than A when dropped.
 
This is pretty simple if you use the http://www.physicsclassroom.com/class/1dkin/u1l6a.cfm" . In particular, if we assume an object starting from a standstill (vi = 0), and using 9.8 m/s2 as acceleration due to gravity then:
t = \sqrt{\frac{d}{4.9 m/s^2}}
Using that formula we can find:
Time to fall 99 meters = 4.495 s
Time to fall 100 meters = 4.518 s
Time to fall last meter (difference) = 0.023 s
Time to fall 1 meter from standstill = 0.452 s

So, it will take almost 20 times longer for the object being released from one meter to hit the ground.
 
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hey thanks dale for putting into formula what I was pretty sure was right
but didnt have the maths skills to articulate it :)

I need to store that formula away somewhere

cheers
Dave
 
Thanks. I was aware that the velocity would be higher for object A, already falling. What struck me was that the object closer to the ground, B, would have a larger initial force of gravity acting on it and that larger force might serve to accelerate object B faster initially. Probably insufficient to overcome the already higher velocity of the falling object 'A'.
 
liometopum said:
the object closer to the ground, B, would have a larger initial force of gravity acting on it

?The force due to gravity is always mg regardles how close the object is on the ground...

to accelerate object B faster initially.

...and the acceleration is always g.
 
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douglis said:
?The force due to gravity is always mg regardles how close the object is on the ground...



...and the acceleration is always g.

No, the OP is right in saying that the value of g differs with height. But at a height of 100 m, the difference is not significant. Even if the the objects would be placed at significant heights, A would still reach first as at its height B is also subject to the same acceleration. However it has already accumulated a certain amount of velocity and any velocity gained by both A and B would be the same from that point. So B's velocity would never exceed that of A.
 

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