Why does a falling object experience constant acceleration due to gravity?

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Falling objects experience constant acceleration due to gravity at 9.8 m/s² until they reach terminal velocity, where air resistance balances gravitational force. Air resistance acts as a retarding force, affecting the overall acceleration of the object. According to Newton's Law, the net force determines acceleration; when gravitational force equals air resistance, the object stops accelerating and moves at a constant speed. The discussion clarifies that while gravity continuously pulls the object down, air resistance increases with speed, ultimately leading to terminal velocity. Understanding these forces is essential to grasping why falling objects behave as they do in an atmosphere.
BRH
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I know that there is constant acceleration due to gravity. A regular force that is applied to a moving object causes the object to accelerate once and as long as the force is present. Meaning, if a ball that is rolling experiences a wind force of 5N the ball's movement will increase 2 mph for example... as long as the wind force is present. Why is it that by gravity the falling object keeps on accelerating every second 9.8m/s. Why doesn't it just accelerate once 9.8 and that's it, until reaching terminal velocity?
 
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How is it that by a falling object if the object doesn't reach terminal velocity, it always falls with acceleration of 9.8m/s. It doesn't matter if the object has a gravity of 5N and air resistance of 4N or if the object has a gravity of 5N and air resistance of 1N... Why doesn't air resistance make a difference as to how much the object will accelerate?
 
BRH said:
I know that there is constant acceleration due to gravity. A regular force that is applied to a moving object causes the object to accelerate once and as long as the force is present. Meaning, if a ball that is rolling experiences a wind force of 5N the ball's movement will increase 2 mph for example... as long as the wind force is present. Why is it that by gravity the falling object keeps on accelerating every second 9.8m/s. Why doesn't it just accelerate once 9.8 and that's it, until reaching terminal velocity?
BRH said:
How is it that by a falling object if the object doesn't reach terminal velocity, it always falls with acceleration of 9.8m/s. It doesn't matter if the object has a gravity of 5N and air resistance of 4N or if the object has a gravity of 5N and air resistance of 1N... Why doesn't air resistance make a difference as to how much the object will accelerate?
Welcome to the PF.

Air resistance does affect the total acceleration of falling objects in an atmosphere. It is a retarding force, The sum of the two forces (gravity down and air resistance up) gives the total force and thus the total acceleration by Newton's Law:

ΣF = ma

:smile:
 
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BRH said:
How is it that by a falling object if the object doesn't reach terminal velocity, it always falls with acceleration of 9.8m/s. It doesn't matter if the object has a gravity of 5N and air resistance of 4N or if the object has a gravity of 5N and air resistance of 1N... Why doesn't air resistance make a difference as to how much the object will accelerate?

berkeman said:
Air resistance does affect the total acceleration of falling objects in an atmosphere. It is a retarding force, The sum of the two forces (gravity down and air resistance up) gives the total force and thus the total acceleration by Newton's Law:

... and when the 2 forces are equal ... the upwards force of the air resistance Vs the downwards force of gravity ...
then the total force = 0 (zero) and with zero force, there is zero acceleration and hence a terminal (fixed ) velocity
Dave
 
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I built a device designed to brake angular velocity which seems to work based on below, i used a flexible shaft that could bow up and down so i could visually see what was happening for the prototypes. If you spin two wheels in opposite directions each with a magnitude of angular momentum L on a rigid shaft (equal magnitude opposite directions), then rotate the shaft at 90 degrees to the momentum vectors at constant angular velocity omega, then the resulting torques oppose each other...

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