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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?

Video (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. [1] [2]

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1 Answer 1

Speed at take-off evaluated:

$105 km/h * 1000 m/km * 1/3600 h/s \approx 29.2 m/s$.

Path length $l_{jump}$:

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

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

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

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

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

Final speed $v_f$:

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

Evaluating $v_f$:

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

Effective landing deceleration component against (normal to) the ground $a_f$:

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

Evaluating bound on $a_f$:

$(4 m/s^2 + 9.8 m/s^2) * \cos[ 35 * \Pi/180 ] \approx 11.3 m/s^2$.

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