How hard something hits the ground

[Shackleford]

Let's say I have a one-kilogram object, and I drop it from heights of 1 meter and 2 meters, respectively, ignoring air resistance/friction and so forth. Are the following statements true?

The objects hit the ground with the same force since the mass and acceleration (due to gravity) do not change.

How "hard" it hits the ground is due to the differing momentum and slight increase in kinetic energy.

 

[dst]

OK, the statement f=ma is not the most accurate (at least, your usage of it in this case). Yes, the mass and acceleration due to gravity are constant. On the other hand, F = ma depends on the resultant acceleration vector on the object, not gravity alone. Newton's original formulation was F = dP/dt, which means change in momentum over change in time. In this case, the change in momentum is ALL of the object's momentum (i.e. F = -mv). Now the impact velocity is given by v = sqrt(2gh) so F = -m * sqrt(2gh). So force of impact for 1m with 1kg is given by F = 1kg*sqrt (2*9.8*1m) = 4.42N. For 2 metres, it is 6.2N. This of course translates into common sense, you will hit the ground harder from a 100ft drop compared to a 10ft drop.

 

[americanforest]

Let's say I have a one-kilogram object, and I drop it from heights of 1 meter and 2 meters, respectively, ignoring air resistance/friction and so forth. Are the following statements true?

The objects hit the ground with the same force since the mass and acceleration (due to gravity) do not change.

How "hard" it hits the ground is due to the differing momentum and slight increase in kinetic energy.

How hard the object hits the ground is dependent on the Power exerted by the ground to stop it, not on the force of gravity (not directly at least).

$$P=\frac{Fd}{t}=\frac{mad}{t}=\frac{md\sqrt{2gh}}{t^2}$$

Where P is the power (ie. How hard it hits), d is the distance to come to a complete stop, t is the time to come to a complete stop, m is the mass of the object, and a is the deceleration of the object. The last equivalence comes from the explanation given by dst above.

 

[Shackleford]

OK, the statement f=ma is not the most accurate (at least, your usage of it in this case). Yes, the mass and acceleration due to gravity are constant. On the other hand, F = ma depends on the resultant acceleration vector on the object, not gravity alone. Newton's original formulation was F = dP/dt, which means change in momentum over change in time. In this case, the change in momentum is ALL of the object's momentum (i.e. F = -mv). Now the impact velocity is given by v = sqrt(2gh) so F = -m * sqrt(2gh). So force of impact for 1m with 1kg is given by F = 1kg*sqrt (2*9.8*1m) = 4.42N. For 2 metres, it is 6.2N. This of course translates into common sense, you will hit the ground harder from a 100ft drop compared to a 10ft drop.

Yes, I know it's the net force on the object. I was aiming for terseness here. I'm in Cal III, so I know all about derivation, blah blah, et cetera. I was hoping some of this was implied. Of course, I know the object dropped from a higher elevation would produce a greater "force." I was trying to figure out how you would express it correctly in physics.

So, the force is given by the product of mass and the instantaneous change in the objects momentum, which would thus translate into the instantaneous change in velocity of the object. At the instant it hits the ground it transfer all of its momentum to the ground and it goes from its calculated velocity to zero.

 

[russ_watters]

The objects hit the ground with the same force since the mass and acceleration (due to gravity) do not change.

Same force as what? Same force as dropped from 1m as from 2m? Definitely not.

When it hits the ground, the acceleration is not the acceleration due to gravity.

How "hard" it hits the ground is due to the differing momentum and slight increase in kinetic energy.

Yes.

 

[Shackleford]

Same force as what? Same force as dropped from 1m as from 2m? Definitely not.

When it hits the ground, the acceleration is not the acceleration due to gravity. Yes.

The same force is the applied gravitational force.

The negative acceleration/deceleration is from the ground.

 

[russ_watters]

Ok, that's what I thought you meant. The answer is no, since when an object hits the ground, it stops quickly. That requires a lot more force.

 

[Shackleford]

Ok, that's what I thought you meant. The answer is no, since when an object hits the ground, it stops quickly. That requires a lot more force.

Right. So, technically, it's incorrect to say how hard something hits the ground. Really, it's how hard the ground hits the object.

 

[mgb_phys]

It's the accelaration ie. how quickly you stop - as they say 'it's not the fall that kills you it's the sudden stop at the bottom'

 

[Shackleford]

It's the accelaration ie. how quickly you stop - as they say 'it's not the fall that kills you it's the sudden stop at the bottom'

Yeah. Ok. Thanks for the clarification, guys.

 

[russ_watters]

Right. So, technically, it's incorrect to say how hard something hits the ground. Really, it's how hard the ground hits the object.

It doesn't really matter how you say it. Each applies a force on the other.

 

[Shackleford]

It doesn't really matter how you say it. Each applies a force on the other.

Yeah, I guess it doesn't matter now that I think about it. Thank you, Newton's Third Law. But the cause of the acceleration is because the object is hitting the ground.

 

[regor60]

OK, the statement f=ma is not the most accurate (at least, your usage of it in this case). Yes, the mass and acceleration due to gravity are constant. On the other hand, F = ma depends on the resultant acceleration vector on the object, not gravity alone. Newton's original formulation was F = dP/dt, which means change in momentum over change in time. In this case, the change in momentum is ALL of the object's momentum (i.e. F = -mv). Now the impact velocity is given by v = sqrt(2gh) so F = -m * sqrt(2gh). So force of impact for 1m with 1kg is given by F = 1kg*sqrt (2*9.8*1m) = 4.42N. For 2 metres, it is 6.2N. This of course translates into common sense, you will hit the ground harder from a 100ft drop compared to a 10ft drop.

You started out right but ended up wrong here. You have dp correct, but forgot the dt part, which isn't knowable, really, without experimentation. The forces above are wrong

 

[dst]

You started out right but ended up wrong here. You have dp correct, but forgot the dt part, which isn't knowable, really, without experimentation. The forces above are wrong

You're correct. I realized that just after I posted, a brain fart I guess. The force comes from the ground acting on the object and then you'll have to take into account loads of things like the depth it penetrates, but I guess that doesn't help. I originally meant to write about impulse but somewhere I got muddled up and wrote force.