Ski Jumper's vertical velocity after 246.5m record?

Question

What would be the vertical velocity of this ski jumper (ski flyer), after he first touches down, after he breaks the record with a 246.5m jump? What g force would he experience as he slows down? http://www.wired.com/playbook/2011/02/worlds-largest-skiflying-hill (the last video)

I think he gets at least 6 seconds of airtime from 26 to 32 second mark.

The ramp gradient at take off is listed at 11 degrees and the take off speed is about 105km/h.

Some statistics are here. http://berkutschi.com/en/front/hills/show/33-vikersundbakken http://berkutschi.com/en/front/news/show/1347-vikersundbakken-approved---worlds-largest-hill

Answer 1

In this solution I have simplified the problem by neglecting air resistance on the ski jumper, and assuming that the hill is dimensionally similar to the one at Vancouver (because I could find more detailed information about the hill at Vancouver here: 1.

We start with the equation $$ v_y^2=v_0^2 + 2a_y(y-y_0).$$ Substituting $a_y=-g=-9.87$ m/s$^2$, $v_0=-v_0 \sin(11.0)=-27.8$ m/s, and $y-y_0=-129.1$ m (which I obtained from increasing the dimensions of the Vancouver hill, based on the assumption that the hills are geometrically similar), we have $$v_y=\sqrt((-27.8^2)+2(-9.87)(-129.1))= -57.6$$meters per second. This would be right without air resistance. Considering the equation $$v=\sqrt(v_x^2+v_y^2)$$we see that for his vertical speed to be 61.4 m/s he would have to be traveling at 114 m/s, or over 250 mph. The fastest recorded free fall speed, done without any special equipment, was 321 mph (speeds like this take significant practice to achieve, but I think a record-breaking ski jumper could pull it off), so the result is plausible. If these assumptions are true, then his vertical speed upon landing would have been 57.6 m/s = 206.6 km/h = 128.4 mph.

This answers the question about vertical velocity.

To answer the question about the g-forces the skier would experience as he decelerates, one would have to know how long it took him to stop. He slows down at a constant rate, because the force due to the kinetic friction of the ice is given by $F=m (\mu g)= ma$, and $a=\mu g$ does not change with time. Thus we could use the equation $$v_f^2=v_i^2+2a(x-x_0)$$ or $$a=\frac{\Delta v}{\Delta t}=\frac{v_f - v_i}{\Delta t}$$to find his acceleration, and then compare this with the value of g to determine the g-force that he experiences, if we knew either the distance or the time that it took him to stop (because we know his initial and final velocities).

Answer 2

Speed at take-off evaluated:

105 km/h * 1000 m/km * 1/3600 h/s =~= 29.2 m/s.

Ansatz for pathlength $l_{jump}$:

$$l_{jump} := (v_0 + \frac{1}{2} a_{e\!f\!f} \, \tau_{air}) \, \tau_{air}$$.

Solving for effective acceleration $a_{e\!f\!f}$:

$$a_{e\!f\!f} := 2 \, l_{jump} / (\tau_{air})^2 - 2 \, v_0 / \tau_{air}$$.

Evaluating $a_{e\!f\!f}$:

2 * 246.5 m / (6 s)^2 - 2 * 29.2 m/s / 6 s =~= 13.7 m/s^2 - 9.7 m/s^2 == 4 m/s^2.

Ansatz for final speed $v_f$:

$$v_f := v_0 + a_{e\!f\!f} \, \tau_{air}$$.

Evaluating $v_f$:

29.2 m/s + 4 m/s^2 * 6 s == 53.2 m/s =~= 192 km/h.

Ansatz for effective landing deceleration component against (normal to) the ground $a_f$:

$$ a_f \le (a_{e\!f\!f} + g) \text{Cos}\bigl( \phi_{impact} \bigr)$$.

Evaluating bound on $a_f$:

(4 m/s^2 + 9.8 m/s^2) * Cos[ 35 * Pi/180 ] =~= 11.3 m/s^2.