# Ski Jumper's vertical velocity after 246.5m record?

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

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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.

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).

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"neglecting air resistance on the ski jumper" I don't think it's a realistic assumption. If I were a ski jumper I would do everything to utilize aerodynamic lift. – Yrogirg Aug 23 '12 at 7:15
@Yrogirg: you would probably be much better off doing everything you could to minimize aerodynamic drag. – Jerry Schirmer Sep 24 '12 at 18:57

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.

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