Hey there! I am from a community called Parkour. We are interested in mastering our environment and surroundings. One thing that we commonly face is whether or not training jumps from a height is reasonable and safe. Some people train and drill jumps from 10-20 feet. Others argue that it should never be done. To explore this, we wanted to explore the physics of the movement. I have done some preliminary calculations but I got stuck on how to convert this to something that can be compared to squatting a weight. The numbers I come up with, shown below, seem very high when compared to those of squatting a 100 kg barbell. Here is what i have so far, I hope someone here can help me... Equation List 1. Force = Mass * Acceleration (F=ma) 2. Velocity = Time * Acceleration (v = at or Vf - Vo = at) 3. Momentum = Mass * Velocity (p = mv) 4. Drop Distance = Initial Velocity * Time to Drop +One-Half * Acceleration * Time to Drop squared (dx = Vo*dt + (1/2)*a*dt^2) 5. Avg Acceleration = Change in Velocity /Change in Time (a = dv/dt) 6. Change in Velocity = Final Velocity -Initial Velocity (dv = Vf-Vo) 7. Change in distance during compression = Average Velocity * Time to stop moving (dx = v*dt) For now lets assume that we are dropping like a formless ball. Cannonballing off of the top of a roof to neglect things like air resistance and the fact that your body will rotate on your center of mass. Let's get to the math. dx = Vo*dt + (1/2)*a*dt^2 --> Assume Vo is 0 since we are dropping from a height into free fall. dx = (1/2)*a*dt^2 -->Rearrange equation 5 to solve for dt. Plug that in to this equation.. dx = (1/2)*a*dv^2/a^2 --> Cancel out the a's and plug in equation 6 which leaves you with this... dx = (1/2)*(Vf-Vo)^2/a --> Once again, Vo is 0, which leaves us with an equation where we have all known parameters and can find our final velocity such that.. dx = 1/2*Vf^2/a --> Rearrange this to solve for Vf (so that we can solve for our momentum later) Vf = sqrt(2*dx*a) OR Vf = (2*dx*a)^.5 -->sqrt = square root. Now we can find out our momentum when we hit the floor as a cannon ball. Lets find out our velocities from different heights.. Sample Calculation at 3 meters (~10 Feet) Vf = (2*3 meters * 9.8 meters/second squared)^.5 = (58.8 meters squared/seconds squared)^.5 = 7.67 meters/second. 1 meter/~3 feet = 4.43 m/s 1.5 meters/~5 Feet = 5.422 m/s 3 Meters/~10 Feet = 7.67 m/s 3.5 Meters/~11.5 feet = 8.28 m/s 6 Meters/~20 Feet = 10.844 m/s 6.5 Meters/~21 feet = 11.29 m/s Now we have some velocities that we will hit the ground with. But how does that translate into the collision/landing. Firstly, the collision will most likely be elastic, since we will not STICK to the earth, but rather bounce back away from it. We can assume that your sneakers or feet will compress by something like 1 millimeter, for calculations sake. From here we must find the time it will take our bodies to stop once we hit the floor (remember this will be pretty fast). We can do this with equation 7. dx = v*dt --> Solve for the time to stop moving dt = dx/v --> Plug in using known values. Sample Calculation using 3 meters/~10 feet dt = .001 meters/(7.67 meters/second)/2 -> we divide by 2 because we are taking the average of 0 m/s (rest) and 7.67 m/s (final velocity) dt = 0.00006520506637 s (6.52 * 10^-5 s or 65 microseconds) to stop moving. Note: 65 us = 65 microseconds 1 meter/~3 feet = 112 us 1.5 meters/~5 Feet = 92 us 3 Meters/~10 Feet = 65 us 3.5 Meters/~11.5 feet = 60 us 6 Meters/~20 Feet = 46 us 6.5 Meters/~21 feet = 44 us Now that we have these times to stop moving once we hit the ground, we need to figure out how much we need to accelerate against gravity in order to support our bodies upright from these heights. Using Equation 5, we can solve for our acceleration. a = v/t Sample Calculation using 3 meters/~10 feet a = (7.66 m/s)/65 microseconds) = 117600 m/s^2 or 12000 Gs (12000 times the acceleration due to gravity. 1 meter/~3 feet = 39200 m/s^2 or 4000 G 1.5 meters/~5 Feet = 58800 m/s^2 or 6000 G 3 Meters/~10 Feet = 117600 m/s^2 or 12000 G 3.5 Meters/~11.5 feet = 137200 m/s^2 or 14000 G 6 Meters/~20 Feet = 235200 m/s^2 or 24000 G 6.5 Meters/~21 feet = 254800 m/s^2 or 26000 G These calculations show that your acceleration will increase linearly, not exponentially. The velocity increases logarithmically, the time decreases slightly exponentially. The gravitational force increases linearly. Maybe I did something wrong...but it seems form these calculations that you can increase your jump hight linearly, or at a constant rate over time. Do note though, that even a 3 foot drop means you are going again 4000 times the acceleration due to gravity. Increasing from a 3 foot drop to a 10 foot drop increases your load threefold, according to this. Unfortunately, thats usually where I get stuck in calculating these things...i even asked some mechanical engineers these things and they got all befuddled too. From what I understand, a 100 kg (220 lb) mass, when resting on your shoulders in push press is exerting 100kg * 9.8 m/s^2 of force on your body downwards. This is 980 Newtons of force. This is really low compared to the Forces and accelerations that we examined during a fall and a 220 lb weight is no joke. To think that a 1 foot drop is harder to execute than a 220 lb squat is pretty absurd.