What height is needed to drop a $72\ \rm{kg}$ mass to impart a $5000\ \rm{g}$ shock onto a test piece I am trying to design a simple drop / shock testing machine.
We want to rate our products to be shock rated to $5000\ \rm g$. 
In order to test our products to $5000\ \rm g$ I plan on dropping a weight onto the product imparting a shock. (See drawing)

My question is;
How high would I need to drop my $72\ \rm{kg}$ weight from the test piece in order to rate by product to survive a $5000\ \rm g$ shock.
I understand that the impulse time and distance is quite critical in the final numbers regarding the calucations. Lets say for calucation sake, that both materials were to be steel.
 A: The following assumes that you want to test the material being impacted. If what you want to test is how well the internals of your object (internal electronics, for example, or the durability of a product during shipping), then this impact test will not give the results you want.
There is no one single height that will get you the shock you require. Let's take this step-by-step.
Your test object must withstand a $5\,000g$ shock, which means it must be able to sustain, at least momentarily, an acceleration ($a_T$, the $T$ referring to the test object) of at least $5\,000g$.
$$a_T > 5\,000g$$
Multiply both sides of the equation by the mass of the test object ($m_T$) gets us the required force to be applied to the test object ($F_T$) by Newton's second law ($F=ma$).
$$m_Ta_{min} > 5\,000m_Tg$$
$$F_T > 5\,000m_Tg$$
By Newton's third law, the force experienced by the test object is equal to the force the falling mass experiences on impact.
$$F_F > 5\,000m_Tg$$
where $F_F$ is the force experienced by the falling mass.
The acceleration of the falling mass can be found with another application of Newton's second law.
$$m_Fa_F > 5\,000m_Tg$$
Here, $m_F$ is the mass of the falling weight ($72\,000\,\textrm{kg}$) and $a_F$ is the acceleration it experiences on impact.
Finally, we can write an expression for the acceleration that the falling weight must experience to put the required shock on the test object.
$$a_F > 5\,000g\left(\frac{m_T}{m_F}\right)$$
You cannot determine ahead of time what the required drop height is because it is different for different materials. Imagine dropping the weight onto a mattress versus a concrete floor. The mattress will experience a much smaller shock because the falling weight decelerates over a much longer distance compared to the concrete floor. You can only find out what shock your test object experienced by measuring the acceleration of the falling weight.
So, the procedure for testing your product is


*

*Attach an accelerometer to the falling weight.

*Drop the weight onto your test object.

*Retrieve the accelerometer data and see if the required acceleration was attained.

*If the acceleration recorded in step 3 was not large enough, then repeat the test with the falling weight starting at a higher starting height or use a larger falling weight.


For step 4, the impact shock should scale as the square root of the drop height. This means that, if you need to double the shock, you need to quadruple the drop height. Tripling the shock requires a nine-times larger drop height. The same goes for the mass of the falling weight: quadrupling the mass results in twice the shock.
A: *

*What's a shock test?


I would have thought that a 5000 g product shock test is to test the ability of what's inside the box to survive rapid deceleration without the internals ripping themselves off their enclosure mounts. Dropping a weight onto a static box won't test this at all if the enclosure doesn't move or flex. Shouldn't you be dropping the product? 


*72 kg weight?


It's not clear you have chosen a 72 kg weight. 


*Required velocity


5000 g is a deceleration of roughly 50,000 m/s². If, for example, this is to be achieved with a 1 mm deflection of the top surface of the product then we can calculate the required impact speed from
$ a = \frac {v^2}{2 s}  $ so for our figures we get $ v = \sqrt {2sa} = \sqrt {2 \times 0.001 \times 50000} = 10 \ \text {m/s} $.
To accelerate the product to 10 m/s by a frictionless gravity drop the height is $ s = \frac {v^2}{2a} = \frac {10^2}{2\times 9.81} = 5.09 \ \text m $.
Note that mass doesn't appear in any of the calculations.

I do not work in this field so maybe I'm missing some key knowledge.
