The tallest mountain in the solar system, Olympus Mons on Mars. It has a height of 25 km, Mount Everest is 'only' 8.8 km tall.

40.2k Upvotes

97% Upvoted

u/[deleted] Oct 08 '22

[deleted]

29

u/The_Original_Gronkie Oct 08 '22

You didn't show your work, so you only get 50% credit.

3

u/Caenwyr Oct 08 '22

That's long enough to scream, run out of breath, inhale again (assuming you wore a suit and tripped over the edge) and keep screaming. Several times that, even.

Heck, it's long enough to make a pretty decent farewell speech!

Assuming the wall is truly vertical, of course. If it tilts back by even one degree, you'll be "bouncing" your way down and you'll have considerably less fun before your meet your inevitable demise. Then again, you'll probably never reach those insane velocities so who knows?
2
u/L-System Oct 08 '22

Uhhh, no? 3.7m/s2 for 30s will only be 1.7 km, not 10...
1
u/thrivingkoala Oct 08 '22

Incredible how easy misinformation spreads… The correct answer (disregarding any air resistance) would be:

s= 1/2at² <=> t = sqrt(2s/a) = sqrt(210000/3,7 m/(m/s²⁾⁾ = 73,5 s

Resulting in a terminal velocity of 73,5 s* 3,7 m/s²⁼ 272 m/s ≈ 980 km/h - with mars’ atmosphere being 0,6% the density of earth’s (equalling the earth’s atmosphere at an altitude of 35 km), this might actually be the attained speed given Felix Baumgartner was at > 1000 km/h at a height of 35 km on his jump. Air resistance seems negligible.

TLDR: r/quityourbullshit u/Tccrdj
1

u/Tccrdj Oct 08 '22

I wasn’t bullshitting, I used online calculators to find the terminal velocity of me at 200lbs falling on mars. Then I took the total KM divided by how fast I would be falling in mph and came up with 28-ish. I can’t remember the speed but it was around 1100mph. So 32,808’ divided by 1100 is 29-ish. Maybe I just did the math wrong and wasn’t bullshitting anyone.
1
u/Tccrdj Oct 08 '22

I reworked my numbers from the way I did it before.

Mass 200lbs Cross section 10sqft Drag coefficient on mars 0.029 Air density of mars 0.02 kg/m3 Gravity on mars 3.721 m/s2 Terminal velocity 4422fps 32808’/4422fps=7.4seconds

So I originally had a few mars numbers wrong. I had the wrong drag coefficient. I think that threw it way off. Feel free to check my math without being a dick.
1
u/thrivingkoala Oct 08 '22
Hey, sorry for my conduct.

The problem with your calculation is that you assume a constant velocity problem at the maximum velocity you'd achieve in free fall, when falling off a cliff is a constant acceleration problem starting at a velocity of zero. Your calculation basically assumes being shot by a canon towards the surface of mars at terminal velocity when really you start out from a speed of zero and only have mars's gravity as acceleration. This means you may reach terminal velocity at some point, but you definitely don't travel all the way at terminal velocity.

Your problem expressed in the movement equations of acceleration a, velocity v and distance s with integrative steps between them is (terminal velocity v_max is your starting speed v_0):
a(t)= 0= const.

v(t)=∫(a)dt= a*t+ v_0= 0*t+ v_max= v_max= const.

s(t)= ∫(v)dt= v*t+ s_0= v_max*t
But the actual problem starts at a speed of zero (v_0= 0) and continues with a constant acceleration a_Mars:
a(t)= a_Mars= const.

v(t)=  ∫(a)dt= a_Mars*t+ v_0= a_Mars*t

s(t)=  ∫(v)dt=  ∫(a_Mars*t)dt=  1/2*a_Mars*t^2+ s_0= 1/2*a_Mars*t^2
Solving the last equation for t with a distance s of 10000 m will give you the correct time t needed to reach mars's surface in free fall.

Edit: formatting
1

u/Tccrdj Oct 08 '22

I get it now. I did think of starting at 0. But honestly I had no idea of figuring out the time difference it would take to go from 0-TV. I was playing around with math and thought it’d be fun to try and figure it out for mars (something I never really think about).