Is Free Fall Acceleration Constant Until An Object Hits the Ground?

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Jynku
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Homework Statement
As a body falls through air starting from rest, its acceleration gets smaller and smaller, eventually approaching zero. Why?
Relevant Equations
free fall acceleration is -9.8m/s^2.
Shouldn't free fall acceleration be constant at -9.8m/s^2 until the object hits the ground and turns zero?
 
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Does this mean in intro physics, when applying linear equation (V = U + at, for instance) on questions like "An object drops from a height of 45m. Find the time it takes for it to reach the ground." It's not taking air resistance into account?
 
Jynku said:
Does this mean in intro physics, when applying linear equation (V = U + at, for instance) on questions like "An object drops from a height of 45m. Find the time it takes for it to reach the ground." It's not taking air resistance into account?
That's correct. There's no air resistance in that equation.
 
Jynku said:
Shouldn't free fall acceleration be constant at -9.8m/s^2 until the object hits the ground and turns zero?
It's not the acceleration that is constant, it's the gravitational force. (And that's only a reasonable approximation when you are close to Earth.)

And the force is still there when you hit the ground.
 
Jynku said:
Does this mean in intro physics, when applying linear equation (V = U + at, for instance) on questions like "An object drops from a height of 45m. Find the time it takes for it to reach the ground." It's not taking air resistance into account?
It is assumed that there is vacuum around the falling object.
Please, see:
https://www.grc.nasa.gov/www/k-12/airplane/termv.html

:)
 
Thanks so much, makes perfect sense!
 
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PeroK said:
:welcome:

Do you know anything about air resistance?
I don’t think there is any serious decrease in gravitational acceleration. We can take it as constant .
 
rudransh verma said:
I don’t think there is any serious decrease in gravitational acceleration. We can take it as constant .
The acceleration is most certainly reduced by air resistance, which is surely the point of the question. Look up "terminal velocity".