What G force will be on an object...?

Question

If an item is dropped from a height of one meter, what will the G force be on the object when it hits the ground?

How you calculate it would be nice.

Answer 1

depends on type of ground and nature of the object, its form, etc...

Answer 2

Use conservation of energy.

The energy initially stored in the object (potential energy) = energy dissipated during impact.  Hint, the form of both sides of the equation is the same except the distances and accelerations are different (mass is the same, of course). Giving you any more wouldn't really help you.

Expect impact forces to be 100's or even 1000's of G's.

Answer 3

Go with the first answer.  The acceleration is constant.

Answer 4

1G. G is the constant gravitational acceleration. The rate of acceleration never changes just the total speed does. You always fall at 9.8 m/s2 unless you take into account air friction and such.

You don't have to calculate constants they are constant. 9.8m/s2 is your answer.

Answer 5

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Answer 6

Considering free fall only, no initial velocity, the speed at impact is given by the conservation of energy:

½mv² = mgh

v = √2gh

v = 4.4272 m/s

The force of impact depends on the deceleration when the object hits the ground, which in turn depends on the time and distance traveled after impact.

Take a 1 kg object, and x = 0.01 m:

I don't remember where I learned this, but somewhere along the line I came across the equation:

2ax = v²

a = ½v² / x

Plugging in v = 4.4272 m/s and x = 0.01 m:

a = ½(4.4272)² / 0.01

a = 980.004992

F = ma

F = 1kg x 980.004992 m/s²

F = 980 N

Here's a page that explains it, along with a handy calculator:

http://hyperphysics.phy-astr.gsu.edu/hba...

(When I first tried using my own equations I kept coming out with twice what they give, then I found my error.)

Edit: What do you think would undergo more force: a 1 kg steel ball falling on frozen ground, or on mud?

To put it another way, it's not the fall that kills you, it's the sudden stop.