How hard do you hit the ground when you fall?

By: Galadriel Billington

1:45PM May 27, 2004

	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   and   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. So--While the horse continues to run on out of your life, you're launched up out of the saddle, AND forwards. 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)