# How to calculate “g's” seen by a dropped object upon impact

How would i go about calculating the amount of "g's" experienced by an object when it hits the ground. I know that force can be calculated somewhat easily with impulse-momentum or work-energy, but what about g's?

Assume in this scenario, that im dropping a 10kg mass from 1m onto a hard rigid surface (like concrete). The object's contact duration is 0.002sec (or 2ms)

It seems that if you use impulse momentum for this problem, mass gets cancelled out but that doesnt make much sense. Im pretty sure that the number of g's experienced by the falling object is a function of mass.

Thanks for everyone's input!

• "I know that force can be calculated somewhat easily with impulse-momentum or work-energy, but what about g's?" Well, it can ... if you have a measurement of estimate of how long the stopping or rebounding takes. – dmckee --- ex-moderator kitten Aug 8 '17 at 23:26

Note, this is a very hard calculation to do in spherical-cow-perfect-physics land. Collisions in such a magical realm are instantaneous, or at the very least have constant forces applied to them during the contact. In the real world, objects compress in all sorts of unintuitive ways.

In your example, you could calculate the acceleration by determining the velocity of the object going into the collision, and the velocity leaving the collision. You know how long the contact occurred for (because you specified it in the problem statement), so you can determine the acceleration it would take to change the velocity by that much. You can then divide that acceleration by the acceleration of gravity to get "gees". This number will be independent of the mass, because you're looking at accelerations, not forces.

In your example, we can use basic kinematics to determine how long the object falls from 1m using $y=\frac{1}{2}at^2+v_0t+y_0$. Plugging our numbers in we get $0 = -\frac{1}{2}9.8t^2 + 1$ which means $t=0.45\text{sec}$. Using $v=at$, we determine that the velocity at impact is $v=-4.4\frac{\text m}{\text s}$. If we assume that all of the energy goes back into the ball, that means that the velocity going back is going to have the same magnitude, but in the opposite direction $v\prime=4.4\frac{\text m}{\text s}$. This means our change in velocity is $\Delta v=8.8\frac{\text m}{\text s}$

Given that the collision happens in 0.002s, the acceleration must be $a=\frac{\Delta v}{t}=4427\frac{\text m}{\text s^2}$. Divide this by 9.8 to convert an acceleration into 'gees' and you get 451.7 gees.

Of course, real objects will be much more complicated than that. They wont always be accelerating at a constant rate through the collision. They also don't always collide elastically.

There is no simple way to do this.

In fact, in the treat-everything-as-a-point-and-surfaces-as-rigid fairly-land where we start learning physics there is no way to do it at all! After all, the thing hits the floor and stops abruptly. You mention finding the force from the impulse-momentum theorem, which is correct if someone tells you how long it takes to stop and you are willing to settle for average force.

The real answer depends on how the body and the surface that it hits deform, and it is quite a difficult problem. The kind of problem that is solved by using large-scale finite element analysis. These days you can do that on a lap top, but when I started out it called for a room-sized supercomputer.

Somewhere in-between is treating the impact as having an ideal spring constant of some kind and using the relationships of simple harmonic motion to deduce a maximum acceleration. Again, though, you need an additional piece of information (the effective spring constant) that is not trivial to guess at.

So let's put some math to this.

• If someone has given the time to stop ($\Delta t$), then you can compute the force of stopping as \begin{align} F_\text{avg} &= \frac{J}{\Delta t}\\ &= \frac{\Delta p}{\Delta t}\\ &= m\frac{\Delta v}{\Delta t}\\ &= m a_\text{avg} \end{align} And "gee" is just a unit of acceleration: $1 \,\mathrm{g} = 9.8 \,\mathrm{m/s^2}$. So you didn't really need the force at all, just the acceleration meaning the impact velocity and the time to stop.

In this point of view the details (including how mass might effect the problem) are hidden in the externally supplied information (how long it took to stop). You can't say "Oh, gees go up with he mass" by any method more subtle than asking for a bunch of those time and then plotting.

• In the effective spring-constant $k_\text{eff}$ point of view, then the maximum acceleration is \begin{align} a_\text{max} &= \omega v_\text{max} \\ &= \omega v_\text{impact} \end{align} where, as usual, $\omega = \sqrt{k_\text{eff}/m}$. Here we can say—assuming the effective spring constant remains the same--how changing the mass changes the peak g-loading: it goes down as the square-root of mass!

Note, however, that the effective spring constant is a complicated function of the material and structure of both the falling object and the surface it lands on. It might not be the same between two objects. Moreover, the whole harmonic assumption can only be expected to work for elastic or nearly elastic impacts and as long as neither object is permanently deformed.

I think your intuition is incorrect...

Given your scenario, you can calculate the velocity of the object after falling 1 m.

You can then calculate the acceleration needed to bring an object from this velocity to zero in the time given, $0.002 \text{ seconds.}$

Finally, you need to add the acceleration of gravity, since the impact with the ground needs to cancel the acceleration of gravity downward, and supply the calculated upward acceleration.

Mass does not enter into the calculation until you try to find the force needed to cause the acceleration.

This calculation assumes a splat, not a bounce...

There is a simple way to estimate this. Calculate the average acceleration when the object hits the ground, divide by $9.8 m/s^2$, and add 1 to account for the weight of the object.

The average acceleration is the change in velocity (= velocity just before impact) divided by the duration of the impact.

Well the g's you are referring to is a ratio of acceleration compared to that of Earth's. For example, if you calculate that an aircraft climbs after takeoff at an acceleration of 49 $\frac{m}{s^{2}}$, then the g's the fighter pilot would experience is $g=\frac{49 \frac{m}{s^{2}}}{9.81 \frac{m}{s^{2}}}= 5$ g