Stopping time of a dropping object

I am having a hard time figuring out how to find the estimated stopping time for both my egg and my craft.

Drop Height (d) = 11.4 m
Mass of the vessel (mc ) = 64.88 g = 0.06488 kg
Mass of the egg (me) = 57.02 g = 0.05702 kg
Average drop time = 1.75 s
(Estimated) Craft stopping time (tc) =
(Estimated) Egg stopping time (te) =

I would appreciate some insight, my brain is fried when it comes to physics. >.<

OK 'Stopping times' will determine the deceleration that occurs
and thus the force exerted on, both vessel and egg.
Remember F=MA and A is delta V / delta T
The quicker the stopping time the higher the acceleration.
The higher the acceleration, the greater the force.

example: In your car:
If you stomp on the brake, you get thrown forward.
If you ease to a stop, you barely feel it.
The same energy is 'stretched out' over a longer time

The purpose of the vessel is to absorb this force by deforming,
and thus extend the stopping time of the egg reducung the force on it,
and keepig it from breaking.

Your question can't be answered from the information you have.

The stopping time will depend on the design of the craft (presumably carrying the egg) .. and the type of surface it collides with.

eg .. a rigid (unyielding) craft meeting a solid (unyielding) surface will bring the whole lot to a rapid stop .. meaning a high rate of deceleration and so large stopping forces will be involved.
On the other hand .. if it collides with a soft cushion .. ..