I recently saw a GroPro video of a monoskier, Trevor Kennison, skiing off Corbet’s Couloir in Jackson Hole Wyoming. After getting over my astonishment that he successfully landed the jump, I wondered how far he dropped. His POV video angle didn’t provide a useful prospective for me to estimate how far he fell. Fortunately, my inner physics geek came to the rescue.

From another video, I determined that Trevor skied off the edge horizontally at a relatively slow speed. Therefore, at the point he went off the edge, his vertical velocity can be estimated at zero. His horizontal velocity doesn’t affect his vertical velocity, but it does determine how far he will fall. In other words, his horizontal velocity determines how far out he will land on the slope. And where he will land, determines how far he will have fallen.

When landing on an incline, your horizontal speed is one factor that determines the length of your fall. The other factor is your vertical velocity. But in this case, Trevor’s vertical velocity was zero, so the length of his fall was completely determined by the speed he went off the edge combined with the steepness of the slope.

I clocked (roughly estimated) Trevor’s fall at two seconds on my computer from his video. The formula required is:

Distance = Initial Velocity + ½ (Acceleration of Gravity)(Time of Fall)(Time of Fall)

= 0 + 1/2)(9.81 m/s)(2 secs)(2 secs)

= 19.62 meters

= 64.37 feet

His vertical velocity at landing can be calculated by the formula:

V = (Acceleration of Gravity)(Time)

= 9/81 m/s(2 seconds)

= 19.62 meters per second

= 70.63 km/hr

= 43.88 MPH

A one second variation either way will make a large change in the fall distance. A one second fall translates into only 15 feet. Where as a three second fall means a fall of 144 feet.

Let’s see that this means for Trevor. Assuming that the incline of the slope is consistent from top to bottom (it’s not), IF his horizontal velocity at the edge was 10 MPH (just guessing here for the sake of argument) and he fell 64 feet, then 5 MPH would create a 15 foot drop and 15 MPH would create a 144 foot drop. Therefore, given this slope, the speed at which Trevor launched was a critical factor in determining the resulting fall length. That means knowing how fast you are going is very important when jumping off cliffs.

### It’s not the fall that kills you, it’s the landing.

If you do a jump and land on a flat surface, the kinetic energy created is determined by:

Kinetic Energy = ½(mass)(vertical velocity(vertical velocity) = ½(m(v)(v)

In this case, 100% of the energy will absorbed by your suspension system, the snow, and your body. If you land on a vertical surface, 0% will need to be absorbed. Hence, as long as you keep falling you are fine (until you land, then you are not fine).

If you land on a 45 degree slope, only 70.7% of the energy will need to be absorbed[1]. Therefore, it is like falling about 2/3 of the distance. A 22.5 degree slope makes it 92.4% of the energy, while a 67.5 degree slope makes it only 38.2% of the energy. This is the reason people are able to land form incredible heights, they land on very steep slopes. But if they overshoot their landing point and land flat, the resulting shock is much greater from a 67.5% to a 0 degree flat landing.

### Shock absorber travel length matters!

The Vertical Force required to dissipate the kinetic energy is determined by:

Work = (Force)(Distance) = Change in KE. (Thanks Ed!) Force = Change in KE/Distance Vertical Force = ½ (mass)(vertical velocity)(vertical velocity) / (compression distance)

= ½ (m)(v)(v)/d

The travel distance of your shock absorber makes a considerable difference in your ability to safely land a jump. All else being equal, twice the travel length of the shock absorber will double the time to cushion the fall, which cuts the vertical force in half. A monoski with 8” of suspension will be able to fall twice as far one with 4” of suspension. When it comes to absorbing impact, time is your friend. You want to do it as slowly as possible. The skiing control problem created by excessively long suspension is a separate issue.

### The landing surface matters too!

The surface that you land on is part of the overall system that absorbs your impact energy. Once the monoski suspension system has been maxed out, it’s either the snow or your body. The same physics applies to the snow that applies to shock absorber travel. The snow compresses as you land on it, slowing you down over time. The longer it takes to become fully compressed, the more the kinetic energy is reduced over time. Twice the compression length creates twice the compression time which cuts the stopping force on your body in half.

Note that six inches of fluffy powder on a solid surface will compress very quickly, while a few inches of dense mushy snow will compress more slowly and thus absorb more impact. Ice barely compresses. Therefore, it’s not just the depth of the snow that matters, its consistency does too.

### Maximizing hang time.

If your goal is to fly in the air for as long and as safe as possible, then you want to launch upward at 45 degree angle and land on a steep declined slope. A 45 degree launch angle will allow you to takeoff at the lowest speed and give you the greatest combination of vertical height and horizontal distance at that level of speed. In other words, in order to get the same hang time when launching at 22.5 degrees, you have to go twice as fast and go horizontally twice as far. Sometimes, speed is your friend and sometimes, speed kills.

Distance = (Velocity)(Time)

Velocity = (Acceleration)(Time)

When you launch upward at an angle, half the time in the air is spent going up and the other half is spent going down (not always, depends on the terrain). This calculation is assuming you are in a not so steep terrain park with a takeoff and landing ramp. If you want two seconds of airtime, then you only fall for one second. Based on our previous calculations, one second of falling time means fifteen feet of fall. Landing on an incline of 45 degrees cuts the impact force to 70.7% of flat making the effective fall force to 10.6 feet.

In order to fall for one second, you need to gain a height of fifteen feet in one second. That means your average vertical speed must be 15 f/s. This requires your vertical takeoff speed (which equals your vertical landing speed) to be 30 f/s = 20 MPH (at the top of the arc your vertical velocity will be 0 f/s). In order to create an average vertical velocity of 20 MPH, you need to takeoff at 40 MPH since ½ the component of your speed is going horizontally and ½ is going vertically at a 45 degree takeoff. Since the ramp is an incline, it will slow you down as you come to the takeoff point. Therefore, you need to hit the ramp at above 40 MPH so that you can takeoff at 40 MPH. The approach speed will be determined by the length of the ramp and the snow conditions on the ramp.

At 20 MPH (30 f/s) of horizontal velocity, you will travel sixty horizontal feet in two seconds.

If your takeoff speed is faster than 40 MPH, you will rise and fall for longer than two seconds. Therefore, you will land further down the ramp at a higher over all speed. Takeoff too fast and you risk overshooting the ramp which will result in loss of impact shock absorption and vastly increased impact force due to your higher fall height.

Note: If you are happy with one second of hang time, then take off at 20 MPH, rise 7.5 feet, and travel 30 feet horizontally. It's a lot safer with much larger room for error.

### Conclusion

The ability to safely jump in a monoski requires far more than just skiing ability. You also need to be able to accurately read the terrain, the condition of the snow, and your takeoff speed, as well as have a monoski with reliable and adequate suspension.

Don’t forget, your body is the final stop when it comes to absorbing the kinetic energy of falling[2]. Like it or not, whatever energy that is not taken by the suspension and snow will be absorbed by you one way or another.

### Notes

[1]. I went over my head here and was saved by my fellow Bates physics classmate and now physics professor, Ed Wiser. Here is what he wrote:

"In the last paragraph of this section, you are describing how the vertical component of the normal force is related to the tangent of the slope. It should be F cos θ.

Slope cos θ

0° (flat) 100%

22.5° 92.4%

45° 70.7%

67.5° 38.2%

90° 0%

As you also write, this would assume that the normal force is the same on all different slopes. But it depends on what the direction the net force is (the acceleration). This would depend on the elasticity of the collision with the ground, and any friction. You can’t really compare this without converting this problem into an impule/momentum problem in 2 dimensions. Then, the time of the impact becomes an issue, which is certainly related to the distance of the compression.

But since your point is that if you launch fast, have a large hang time, then the speed at impact grows as the square of the time you fall, then… you got all of that correct."

[2]. Here is a handy splat calculator for determining some of the physics of falling.

## Comments