I was motivated to do the Math. So here it is, I think... if it was a 7m gap and a, 11m drop, that gives us a hypotenuse of 13.something, which I rounded up to 13.1m to roughly account for their curved travel path. That is 42.9790 feet of travel. h(t) = 42.9790 - 16t2 for finding the amount of time t, traveled in the air. If h(t) is the height of impact, 0, then that gives us a travel time of 1.63896 seconds. Solving for velocity = gravity x time, he was ~roughly~ traveling at about 17.88 mph when he hit the pads. If someone wants to do the weight that was put on his knees at the point of impact, I would like that.
Horizontal and vertical forces are independent of each other. I'm going to assume his mass is about 170, so about 76.5kg. Also keep in mind this is in a frictionless environment.
P
e = mgh = 6747.3Joules of Energy
K
e = 1/2mv
2 solving for velocity = 13.28 m/s ~ 40.48f/s ~ 29.17mph at the time of impact.
p = mv = 1015.92 kg*m/s
J = F
nett solving for time = 1.35seconds in the air
Now here is where the numbers get a little iffy. There is no way to accurately measure the time from impact to stop without being there with very expensive equipment, so I'm going to guess it took about .2seconds to stop, which I feel is VERY gracious, as it was probably much quicker.
a= ( v
f - v
i ) / time = decelerating at a rate of 66.4 m/s
2F
net=ma = 5079.6 N which is equivalent to squatting about 518.32kg or 1142.69lbs.
This is interesting but doesn't relate to the problem directly:
P = W/t = 33736.5 Watts of power, or 45.24 horsepower, enough to run a standard boat.
