[RiversEdge]

Before jumping, the guy trying to find the wormhole should have employed a camera operator with a higher def video recorder. Did the best I could w/ what was dealt to me.

[REVZZYBAN]

Ouch!! On the side, I bet that hurt. Good bounce though. The rescue brigade was warned well in advance of the free fall. They scattered like roaches.

[kellyhound]

lucky he didn't hit the rescuers

[Suicide_Note]

A millisecond of agony, and then it's all over.

[RedOnes]

Pretty sure you can see contents of his skull spill out for a brief moment on impact.

[SolonJhee]

Yea we don’t want a repeat of that dude who got suicide jumped on.... Bad way to go i say.

[pokerplay]

All they needed to do was put down a tall glass of water for him to aim for.

[docfr8]

7.5
He be like "fuck you, I'm going on my terms.... "

[Humanation]

That's not in Hong Kong. You can tell from the blue carplates.

[b1gpoppa]

All I can think about when I see this is Bernadette from the Big Bang Theory explaining the science behind the free falling - and polishing my purple-headed junket-pumper till I cover her glasses with my essence.

This case, which applies to skydivers, parachutists or any body of mass,
m {\displaystyle m}
, and cross-sectional area,
A {\displaystyle A}
, with Reynolds number well above the critical Reynolds number, so that the air resistance is proportional to the square of the fall velocity,
v {\displaystyle v}
, has an equation of motion
m d v d t = m g − 1 2 ρ C D A v 2 , {\displaystyle m{\frac {\mathrm {d} v}{\mathrm {d} t}}=mg-{\frac {1}{2}}\rho C_{\mathrm {D} }Av^{2}\,,}

where
ρ {\displaystyle \rho }
is the air density and
C D {\displaystyle C_{\mathrm {D} }}
is the drag coefficient, assumed to be constant although in general it will depend on the Reynolds number.
Assuming an object falling from rest and no change in air density with altitude, the solution is:
v ( t ) = v ∞ tanh ⁡ ( g t v ∞ ) , {\displaystyle v(t)=v_{\infty }\tanh \left({\frac {gt}{v_{\infty }}}\right),}

where the terminal speed is given by
v ∞ = 2 m g ρ C D A . {\displaystyle v_{\infty }={\sqrt {\frac {2mg}{\rho C_{D}A}}}\,.}

The object's speed versus time can be integrated over time to find the vertical position as a function of time:
y = y 0 − v ∞ 2 g ln ⁡ cosh ⁡ ( g t v ∞ ) . {\displaystyle y=y_{0}-{\frac {v_{\infty }^{2}}{g}}\ln \cosh \left({\frac {gt}{v_{\infty }}}\right).}

Using the figure of 56 m/s for the terminal velocity of a human, one finds that after 10 seconds he will have fallen 348 metres and attained 94% of terminal velocity, and after 12 seconds he will have fallen 455 metres and will have attained 97% of terminal velocity. However, when the air density cannot be assumed to be constant, such as for objects falling from high altitude, the equation of motion becomes much more difficult to solve analytically and a numerical simulation of the motion is usually necessary. The figure shows the forces acting on meteoroids falling through the Earth's upper atmosphere. HALO jumps, including Joe Kittinger's and Felix Baumgartner's record jumps also belong in this category.