# Physics Paradox about Newtons Second Law $F=ma$

I randomly thought of this- I'm surprised it took me so long to think of it, and I'm SURE I'm making a simple oversight of something completely trivial, but here it goes. (Im only in high school taking normal physics, so don't laugh if this is stupid.)

1. If force equals mass times acceleration, wouldn't a basketball dropped from the top of the Eiffel tower exert the same force on the ground as a basketball dropped a foot off the ground? They both have the same mass, and they both are accelerating towards the ground at a rate of g=9.81 m/s^2. (I don't know what terminal velocity is that well as im only in physics 1 in highschool, but just assume that air drag is not important and the ball doesn't reach terminal velocity.)

2. Also, if a ball IS dropped high enough to reach terminal velocity, then it accelerates at 0 m/s^2, so it has a force of ZERO when it hits the ground?

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While the ball is in free fall, yes, the net force it exerts on the ground is roughly independent of its height above the ground, and is independent of its speed.

However, you are interested in the force between the basketball and the ground while the ball is colliding with the ground, and that is entirely a different matter.

When the ball is colliding with the ground, its speed drops from a very high speed going down to a speed of zero, or to a speed going up if it bounces. That means that the acceleration of the ball during the collision, points up. While the ball is falling, its acceleration is down, and that acceleration changes directions during the impact.

Further, since the duration of the impact is very short, the acceleration is extremely high. If you drop the ball from a high height, it will be going very fast when it strikes the ground, and this also means the acceleration is very great. Thus, a ball dropped from higher exerts the same force on the Earth while falling, but a higher force during its impact because its acceleration is higher during impact.

If a ball falls at terminal velocity, there are two significant forces on it: gravity and air drag. The force from gravity is unchanged. Gravity still pulls the ball down and the ball still exerts and equal and opposite upward force on the Earth. However, the force from air drag pushes the ball up, and the equal and opposite force from the ball on the air is down. Overall, there is zero net force on the ball. If we consider the atmosphere and Earth excluding the ball as one system, there is zero net force from the ball on that system as well.

When a ball moving at terminal velocity hits the ground, it will suddenly pick up a huge acceleration that it didn't have a moment before. This means there will be a big net force on it that wasn't there a moment before. The fact that the net force was zero before the impact does not mean the impact itself occurs with zero forces at play.

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I think the other two answers may have overlooked the source of your confusion, which is quite simple.

The $F$ in $F=ma$ is the force being exerted on the object of mass $m$ to give it the acceleration $a$, not the force that that object will exert when it hits something.

In the case of your example, the force of gravity on the basketball is independent of the height. The force that the basketball exerts on the ground is an entirely different matter.

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This is the case of jerk in physics , rightly pointed above, so when the ball hits ground it change in acceleration is 2a in very less time. This is the reason why you get more injured when you fall from high tower.

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The problem is that 9.81 m/s^2 is the acceleration during the fall, not the impact.

Let's suppose the impact lasts 0.1 s when a ball hits the ground at 10 m/s, assuming inelastic collision (meaning, the ball doesn't bounce back), average acceleration DURING impact will be 10/0.1 = 100 m/s^2 rather than 9.81 m/s^2 and that's the acceleration you should be using for the formula

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