G-Force of Dropping a Stone and Football from 1.5m

[disregardthat]

Hello, I am new here, and finally i found a forum where i can ask all the things i have wondered (and discussed) about! I do not know so much about physics laws and such, but i find it very interesting. I have a few question i really want answer to. Well, i will ask one here now:

How many G's will a stone experience if you drop it from 1,5 meter height? Can you compare it to what a football will experience from the same height? (this is a part of a discussion i have had with one of my friends)

 

[dav2008]

Hello.

A g is a unit of acceleration. (Note that it is written in lowercase). It is defined as being 9.80665 m/s². That definition comes from the acceleration that an object experiences due to gravity at about the Earth's surface. If something accelerating at 9.8 m/s² that means that every second its speed increases by 9.8 m/s.

That being said, if you drop the football and stone from a height that's near the surface of the Earth, they will accelerate downwards at approximately g (ie they're experiencing 1 g). When you factor in air resistance/buoyancy, the football would accelerate slower than your typical rock but if you drop it from 1.5 meters then the effect would be almost negligible.

Now if you are asking the acceleration that each object undergoes when they strike the ground then that's a different story. An object dropped from 1.5 meters on the Earth will be traveling at $\sqrt{2(1.5)(9.8)}m/s$ or 5.4 m/s when it hits the ground. The acceleration that it undergoes depends on how soft/hard the object is and how soft/hard the ground is, among other things. You can safely say that the stone would experience more gs as it hits the ground because it is unable to deform like a football.

I can't really give you exact numbers but I'm sure someone can look up the numbers in a chart for the given materials and such. My estimate would be somewhere around 10g, give or take.

 

[disregardthat]

Thank then i believe i won the discussion then
But what if you drop something with extreme hardness. like something a thousand times harder than a diamond. would the g's stretch out as the hardness of the object? if a stone has a hardness of 20 (no measurement, just hypothetical), gets 10 g. will then a objekt with a hardness of 40, get 20 g? and a object of 1000, get 500 g? just wondering where the line is. (sorry, but i am no good at the english physical expressions)

 

[berkeman]

The term for what you are asking about is "impulse". It is a measure of how hard the shock is of hitting something. I googled impulse force tutorial, and got lots of good hits (pardon the pun). Here's one:

www.freestudy.co.uk/dynamics/ impulse%20and%20momentum.pdf

The harder and harder you make the two objects that are colliding, the higher the peak of the impulse (assuming that they don't shatter on impact).

 

[leright]

The term for what you are asking about is "impulse". It is a measure of how hard the shock is of hitting something. I googled impulse force tutorial, and got lots of good hits (pardon the pun). Here's one:

www.freestudy.co.uk/dynamics/ impulse%20and%20momentum.pdf

The harder and harder you make the two objects that are colliding, the higher the peak of the impulse (assuming that they don't shatter on impact).

The impulse isn't quite the measure of the "shock" of hitting something since, for instance, 20 F*t can be spread over a large time interval or a very small time interval, and over a very small time interval the shock is very high for a given impulse, but over a very large time interval the shock is very low for a given impulse. Therefore, impulse alone is not sufficient to determine shock.

However, you are correct in saying the high point in the force-time curve determines the shock. I wouldn't say the peak in the impulse is the maximum shock though, since the impulse is the integral of the force-time curve and the maximum impulse in the impulse-time curve is actually where the force has died down (all of the area of the force time curve has been integrated over.)