How does air resistance affect falling objects of different weights?

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Air resistance significantly impacts the fall time of objects with different weights and shapes. In a vacuum, all objects fall at the same rate regardless of mass, but in real-world scenarios, air resistance alters this outcome. Heavier objects may not necessarily hit the ground first; their shape plays a crucial role in determining fall time. For example, a bowling ball with a parachute can fall slower than a lighter pebble. To calculate fall time in a vacuum, the formula t = √(2d/a) can be used, but adjustments are needed for air resistance in practical applications.
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gravity and air friction?

hey guys i am little confused about how two objects will react each having different weight when it is allowed to fall from a building of 20 floors...for eg if 1 is weighing 100 kg and other one 45 kg so 100 kg will touch the ground first but how much time will an object weighing 50kg take to touch the ground from a height of 300 feet.and consider object's height to be around 6 feet.
 
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Actually that is incorrect to say that the heavier object will hit the ground first.

If we consider that you drop them in a vacuum (and the base of the objects are at the same height, so size of objects don't matter)
then 2 objects will hit the ground at the exact same time regardless of their mass.
So yes, in vacuum a bowling ball and a feather will fall at the same rate.

Now.. if we try to consider more realistic scenarios, you will have air resistance. But you still cannot say that a heavier object will hit the ground first, it completely depends on the shape of an object.
So a bowling ball obviously hits the ground before a feather.
But what about a bowling ball strapped to a parachute compared to a pebble?
The pebble is lighter but hits the ground before the bowling ball+parachute.So in the ideal situation (in vacuum) all objects fall at the same rate (near earth) regardless of size and shape. The only thing that matters is the height you drop them from.
 


spacelike said:
Actually that is incorrect to say that the heavier object will hit the ground first.

If we consider that you drop them in a vacuum (and the base of the objects are at the same height, so size of objects don't matter)
then 2 objects will hit the ground at the exact same time regardless of their mass.
So yes, in vacuum a bowling ball and a feather will fall at the same rate.

Now.. if we try to consider more realistic scenarios, you will have air resistance. But you still cannot say that a heavier object will hit the ground first, it completely depends on the shape of an object.
So a bowling ball obviously hits the ground before a feather.
But what about a bowling ball strapped to a parachute compared to a pebble?
The pebble is lighter but hits the ground before the bowling ball+parachute.


So in the ideal situation (in vacuum) all objects fall at the same rate (near earth) regardless of size and shape. The only thing that matters is the height you drop them from.

i completely agree with your point.but what if i want to know in how much time a normal person will reach the ground from a height of 270 feet..can you just calculate or tell any formula to calculate that?
 


You can do that, it's actually quite simple if you assume they are falling in a vacuum (zero air resistance)

t=\sqrt{\frac{2d}{a}}
where a=9.8m/s^{2} make sure to use consistent units, so if you want to use d=270ft you'd have to convert this to meters, or convert a into feet per second squared.

If you want me to show you from where we derive that equation just ask.

Also, if you were asking for how to find the time if we DO have air resistance, then there's a bit more to it.

Sorry for the short answers I have to take care of something right now. But I think I could help you if you are interested in the air resistance problem.
 


Using that formula t = root 2D/a i was able to achieve time that is 4.09 seconds but that is when only gravity is affecting but in real world there is some air resistance so how can we find that and adjust in above formula
 

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