How do i calculate the force of which this falling object will hit the ground

So im planning on launching a weather balloon with an HD camera. Im attaching a parachute but i would like to consider the worst case scenario in which the parachute doesn't open, thus allowing my camera box to hit the ground at full force. My question is what is the force? Lets say the object weighs 3 lbs (1360 g) and falls from 100,000 ft. I'm not sure if it can be calculated from this information but its all i have to work with. thanks!

Worst worst case, your camera experiences free fall (no friction). From h = 100,000 ft (30,000 m), the speed v with which it hits the ground (approximating also a constant acceleration g) follows from the standard equation

v = sqrt( 2 g h)

The momentum it has then is m v, so

momentum p = m sqrt(2 g h)

This momentum is reduced to momentum zero by the force the ground exerts on the camera.
Now suppose the time between the camera first touching the surface and (without bouncing) coming to rest is dt, then from Newton's F = dp/dt we can calculate the average force F:

Suppose dt = 0.2 seconds, then

F = 1.260 kg * sqrt( 2 * 9.81 m/s^2 * 30000 m) / (0.2 s) = 4833 N .

This will most definitely ruin your camera.

Now, air drag will reduce the speed to an ultimate velocity (the so-called terminal speed).
An estimate for this terminal speed for a relatively compact object like a camera with a failing chute might be about 250 km/h .
That's about 55 m/s .
If you put that in the formula you get

F = 434 N (which is the weight of 44 kg ). So if your camera can withstand that weight on it, this analysis suggests it can withstand a fall.

So, I've had to make a few assumptions; they are as follows:

C_d [drag coefficient] = 0.5 (a "rough" sphere)
rho [air density] = 1.225 kg/m^3
A [area of camera face] = 48 in^2 = 0.31 m^2

These values allow us to calculate the terminal velocity.