Unfortunately the only useful formula you have to work with, likely you are going to need to derive your drag coefficient from observation. Your problem is a bit of a chicken and egg kind of thing.
For instance you can observe what the terminal velocity becomes through careful testing and from...
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Your method is not going to be useful if you don't work in consistent units. So lose the inches. And keep everything in Standard Units - meters, kilograms, seconds ... because these are the units of gravity, etc, that you want to use.
Secondly with the terminal velocity your...
Since it is to be a complete transfer of energy, then you know the energy directly from the height the pendulum is released and where it ends up.
Since this energy is 1/2 mv2 of the block after impact then you know velocity, and from that it's just like horizontally launching a block from a...
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Well what happens at impact?
You have a collision of some sort, and the pendulum regains a specific height below the original release point. As you show this gives you a change in energy, but I think you have calculated the wrong change. The pendulum starts at .73m above and...
You will note that to conserve both energy and momentum after impact with twice the mass involved, something has to give.
You at once have Va = Va' + Vb' and Va2 = Va'2 + Vb'2.
Since Vb' must be non-zero, there is only one condition that can eliminate the extra middle term 2Va'Vb', namely...
This is where it all gets to be a mess.
If that is the order, then she must make quiet whoopy with the husband, but all the delivery lads in the neighborhood ... well those would be horses of a different color.
I think they want you to accelerate that from rest. So ...
W = q*ΔV = 1/2*m*v2
Which means that the ratio of v2 = 2:1, since v2 is proportional to the charge.
Google does it nicely. That was why I gave you the link.
As to why ...? Well, it's because it is the inverse function that yields the angle when you have the sides and want the angle.
Learn it, as this likely won't be the last time you may need to use it.
Depending on the cross sectional area of the bouquet and the amount of lace frills and ribbons adding drag, I'd guess the terminal velocity of a bridal bouquet at something less than a Tim Wakefield change-up, or maybe about 60 - 70 mph.
Dropping rice (not in a bag of course) might be more fun...
Without telling you the step, they have used the arc tan of 4.818 to yield the angle in degrees.
http://www.google.com/#hl=en&q=arctan+4.818+in+degrees&aq=f&oq=&aqi=&fp=KxYPMM6r3XA