Do Dropped Objects Reach the Ground with Equal Force?

[Rasine]

If two objects of equal mass are dropped...one from h1 and the other fro h2...will they hit the ground with the same force?

Because the only force acting on the objects is gravity so we can say:

F=ma

and the acceleration is constant...so the forces must be the same.

Is this correct? Intuitively, this seems not to be so.

 

[Xezlec]

First of all, "hit with the same force" doesn't really mean anything. Things don't hit with "force", by the physics definition of the word force.

What you are saying is since the two objects accelerate at the same rate and have the same mass, then the same amount of force must be pulling on both of them. You are correct, they are both pulled toward the ground with the same force, specifically, their weight. All this says is that an object's weight is proportional to its mass regardless of where it is located.

 

[Xezlec]

Let me add something to my last answer: you might also be asking why an object that has been falling for a certain distance exerts a certain amount of (average) force on the ground during a collision. Let me try and write out the chain of events for you:

stage 1:
A 2 kilogram object is being held 4.9 meters above the ground. it is then released.

stage 2:
The object falls, accelerating downward at 9.8 meters per second per second. Of course, for an object with a mass of m, an acceleration of a is produced by a force of F = ma. This means that something must be pulling downward on the object with a force of 9.8 x 2 = 19.6 Newtons. That something is gravity.

stage 3:
In 1 second (do the math if you don't believe me), the object reaches the ground. At this point its velocity is 9.8 meters per second (again, do the math). Now, the ground resists the object's further motion. In other words, the ground acts to reduce the object's downward velocity to zero very quickly.

Changing a velocity is acceleration, of course. Specifically, the ground changes the velocity from 9.8 meters per second to zero in a short time -- let's just say 0.05 seconds. That means an average acceleration of (0 - 9.8) / 0.05 = -196 meters per second per second. (Note that this is only the average acceleration here. The acceleration actually changes a lot during that 0.05 seconds)

So, the average force exerted by the ground on the object during that 0.05 seconds must be mass times acceleration, which is 2 x (-196) = -392 Newtons. That's a lot. And of course the exact opposite, +392 Newtons, must be exerted by the object on the ground (because every force has an equal and opposite reaction force).

So, you see, the long amount of time (1 s) that gravity spends pulling on the object builds up enough velocity that if the ground (or some other unfortunate thing) tries to get rid of all that velocity in a very short time (0.05 s), it's going to need a lot more force.

 

[gamesguru]

The way a force is exerted on a falling body when it hits the ground is actually quite complicated, and not completely understood because of the function of time that the force obeys.
Assuming the decelerating force with the ground is constant (which is fairly accurate),
$$W=Fd=\frac{1}{2}mv^2$$.
If we divide by d, which is the distance the object takes to stop when it hits the ground, we see a relationship where the force increases as d decreases and v increases (the height increases):
$$F=\frac{mv^2}{2d}$$.

We know that as h increases, v will increase. So if they both discharge over the same distance d, the one dropped from higher exerts a larger force. You can see this directly if you recognize that $\frac{1}{2}mv^2=mgh$, thus:

$$F_{exerted}=\frac{mgh}{d}$$.

Hope that makes it more clear