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This one's just for fun...
How much harder do you hit the ground when bucked off, if:
your horse is galloping instead of standing still? Come to think of it, what if you just fall right off? Or if the horse is galloping and you just slip off, is it a harder fall than getting bucked off standing still?

	Okay, let's consider the question. Say you're a rider of mass m on a horse who is h tall, and he bucks with a force of F. To reference the equations I'm using, go here: Constant-acceleration equations. If you're not interested in how I found it, just in the answer, then skip to the bottom ;)
	The force F imparts a upwards velocity vbuck (that is, you are suddenly flying upwards). You go upwards more and more slowly, until gravity (g) has halted you at a height h2 above the saddle. Then you go back down. You are essentially falling from a standstill at a height d. F=ma   -->   a=(vbuck-0)/tbuck F=m(vbuck/tbuck)   -->   vbuck = (F tbuck)/m 0=vbuck² + 2g h2   -->   h2=(vbuck)²/2g so: You fall a distance d=h+h2.
	Okay, now suppose you are on a horse who is merrily galloping along at a speed v. This horse now bucks with a force F. For convenience, I am going to assume that air resistance doesn't make a difference in the short distance you're falling (it won't really), so v doesn't change.
	The force F imparts the same upwards velocity vbuck. However, instead of flying straight up, and then falling from a standstill, you go up the same distance...you just also go forwards a distance xup. Then, while you are falling from a height d, you are continuing to move forward a distance xdown. When you eventually hit the ground, you are falling down at the same speed as if the horse had been standing still...you are also falling *forward* at the speed the horse was running. You still fall a distance d=h+h2.
	Okay, so this is the difference in the fall: one goes straight down, one goes forward and down. The height of the fall is the same--the time of the fall is the same. d=g (tfall)²/2 tfall = sqrt(2d/g)

Okay, I've tried to keep this uncomplicated so far, but I'm going to make it into a 2 dimensional problem now--I apologize ;)
vy_impact is the same in both situations; it's entirely dependent upon the height of the fall.
vy_impact=g t_fall

vx_impact for the standing buck is 0; for the galloping buck, it is v.

v_impact = sqrt [(vx_impact)² + (vy_impact)²]
When vx=0, then v_impact=vy.

The force hitting the ground is again F=ma. In this case, the acceleration is:
a_impact=(0-v_impact)/t_impact

F_impact=ma   -->   F=m(-v_impact/t_impact)

If you skipped to the bottom for the answer, here it is!
You can play along using the Javascript form at the bottom of the page.

Okay, the only difference is in the "v_impact" term; can that really make a huge difference? Let's check. We have: m - weight of the person...say, 100 lb
h - height of the horse...say, 15.2 (5.2 ft)
v - speed the horse was going...initially, let's say he's standing still
F_buck - force of the buck...say, 100 pounds--enough to toss the rider 4 ft up above the horse's back
t_buck - time the buck took (I'm assuming 1/2 second)
g - gravity (32 ft/s²)

I run it through a short Matlab program, and...
The force of landing is 151 pounds.

Now, let's say he was going 15 mi/hr (22 ft/s):
The force of landing is 205 pounds.

Yes, it makes a difference!

But wait! Suppose that a running horse can't buck as hard as one standing still. Say, only half as hard (F_buck=50 lb -- only enough to toss the rider up 1 foot):
The force of landing is... 185 pounds.

Okay, now say, you simply fall off the galloping horse (no bucking). So, say he was running all out and hit maybe 30 mi/hr (44 ft/s), and then F_buck=0.
Now the force of landing is: 298 pounds!

How does this compare to simply falling off a still horse? (aside from the embarassment factor):
Force of landing: 114 pounds.



How bad was YOUR worst fall?

	Weight of Rider (in pounds)
	Height of Horse (in hands, or percentages of hands; ie, 15:2 would be 15.5)
	Speed of Horse (in ft/s) -- Use "0" for a standing buck or a simple fall.
	Height Tossed Above Horse (due to buck) (ft) -- Use "0" for a simple fall
	Velocity at Impact (ft/s)
	Force of Impact (lb)

If you need to convert numbers, here's a good place to do it:
http://www.onlineconversion.com/



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