# Does a falling object hit the ground with a force of its mass times a of gravity?

Does a falling object hit the ground with a force of its mass times "a" of gravity?

Hi,

When an object is falling through the air and in free fall with no air resistance, the applied force is the net force on the object: the object's mass times the acceleration due to gravity (or -9.8m/s^2). I am wondering if all objects hit the ground with the force of gravitational attraction? If so, then the interraction with the pavement must determine how high the ball bounces. I am learning about momentum and all the collisions are in terms of an object's mass times velocity. I know this is derived from Newton's second law, but I want to look at it terms of f=ma.

What is the initial force that an object hits the ground with? Is this the gravitational force of attraction or m times acceleration due to gravity?

What is the initial force that an object hits the ground with? Is this the gravitational force of attraction or m times acceleration due to gravity?
Force = rate of change of linear momentum

The best approach to understanding force in collisions is to use:
Force = rate of change of momentum
ie Force = (change in momentum)/ (time taken)
How quickly an object comes to rest depends on the nature of the surfaces and any 'give' in the colliding objects.
Just read grzz response .... I agree

So when a falling object collides with the ground is the force the ground exerts up on the object not constant then since the ground gives?

I am simply trying to understand the forces in a free body diagram of the ball in different stages of the collision. The ball strikes the ground with a certain force? What is that force?( I though it was the object's weight)
The ground provides an equal and opposite reaction to slow the ball. Then how does the ball bounce back up? Where is that extra force coming from?

I really appreciate the help, but can someone please answer in a detailed manner? I know you are busy but you will really help someone who has put a lot of time into this question.

An important concept they mention is Impulse, Force * time = change in momentum

So for an object with a mass M and velocity v, the momentum is Mv, The force that object applied when contacting the ground depends on how quickly it's momentum is changed. It's mass won't be changing, so it's velocity will. So the force it applied depends on the rate of change in velocity, also known as acceleration. This brings us right back to F = ma, in this case the a is not the acceleration of gravity, but the deceleration of the object as it hits the ground.

* also, bouncing is from the elasticity in the ball, not "extra-force" from the ground, as the bottom of the ball hits the ground and stops, the top keeps coming, pushing out the sides, which then spring back in, causing pressure on the ground, giving the ball unbalanced forces and upward acceleration? Is this a good description?

And the force the ball applies to the ground is what the ground applies up on the ball to slow it down, right? But then how does the ground push the ball back into the air? The ground has already provided its reactionary force to slow down the ball to a velocity of zero?

I think you were typing your response when I added the other post.

So the elastic forces created as the ball compresses must propel the ball to its original height right? I do not think the pavement applies a constant force, so when the ball bounces back, is the force applied to the ball as it leaves the pavement the same as when it just contacted the pavement?

rcgldr
Homework Helper

So the elastic forces created as the ball compresses must propel the ball to its original height right?
Only if the collsion were completely elastic, with no loss of energy in either the ball or the ground. This would not happen in a real world situation, as some of the energy will be converted into heat and/or permanent deformation of the ball and/or ground.

As mentioned above, the amount of force involved depends on the rate of deceleration, which is related to the stiffness (the ratio of force versus deformation) of the ball and ground. The force involved with a billard ball, which is very stiff, will be much greater than the force involved with a rubber ball, which isn't as stiff as a billiard ball, assuming the ground is at least as stiff as the billiard ball.

Thank you! Last comment: if it was a completely elastic collision, no loss of energy, and pavement does NOT give at all (the ball simply hits and bounces), then would the ground decelerate the ball with the force it applied which since there is no give or deformation would this be the object's mass times acceleration due to gravity?

Lets use the example of a 1kg ball dropped from 2m, the force of gravity is 9.8N, The ball's KE after 2m is to 19.6J, which means a velocity of 6.26m/s and a momentum of 6.26kg*m/s

If the ground applied 9.8N to the ball, it would take .64 seconds for the ground to stop the ball, that is a LONG time...(equal to the time it dropped actually) Most times when I see something bounce it is rather quick, like .1s or less I would think, if it did take .1s the force would have had to have been 62.6N, much greater than gravities pull on the object.

rcgldr
Homework Helper

If it was a completely elastic collision, no loss of energy, and pavement does not give at all (the ball simply hits and bounces), then would the ground decelerate the ball with the force it applied which since there is no give or deformation would this be the object's mass times acceleration due to gravity?
No. The force would be related to the rate of deceleration of the ball, which would be much greater than 1 g of deceleration. The amount of peak force would be much greater than the weight of the ball. Also, the ground is going to give somewhat, but in an ideal situation, the ground's response would also be elastic, and return all the energy back to the ball.