small math problem

[ecookman]

If you were to jump off a cliff facing north and a easternly wind of 5km/h was present during the jump how fast would you fall and where yould your landing point be if the cliff was 100km high... (assuming if there was no wind you would go straight down)


How would I go about solving something like this lol

[Sniper4Life]

umm...are you sure that you read the question properly?

like instead of easternly wind of 5 km/h
it could be an easternly wind MOVING YOU at 5 km/h...cus with what you wrote...i dont see a solution O.O

clearly that question overlooks many things?

[ecookman]

well i know there would be some physics calculations to determine how much the wind is pushing you while falling...but uea that is kinda of a brain blower...ill just stick with the wind moving you at a constant eastern direction of 5KM/h


p.s for all of you wondering this is something i thought of...and was wondering how to solve it..no school involved

[Tony]

ecookman @ Sat May 09, 2009 8:56 pm wrote:

if the cliff was 100km high...

Mnt. Everest is 8.8 km from the sea level.... Passenger planes don't fly above 20 km... just saying.

Those are typical high-school level assumptions -- 100% force transfers and 0% friction, no concept of terminal velocity... Those usually limit the problems to application of one or two simple formulas.

@ecookman -- your hint is that the wind has no impact on the downwards fall, so that would take the same amount of time as with no wind.

[endless]

Tony @ Sat May 09, 2009 9:21 pm wrote:

ecookman @ Sat May 09, 2009 8:56 pm wrote:

if the cliff was 100km high...

@ecookman -- your hint is that the wind has no impact on the downwards fall, so that would take the same amount of time as with no wind.

this is true.
and if you want a remotely realistic time (even through your height isn't very realistic) air resistance would be a huge factor, and even altitude.

although i believe this is beside the point you are trying to make. if you look at this as a triangle, the resultant vector of the triangle being the hypotenuse would be equal to it's adjacent, meaning no sideways movement.

[kyleisgreat]

Yeah it really depends how accurate you want to make this. If you can make the following assumptions:
1. The force of gravity is always the same (9.8m/s^2)
2. There is no wind resistance
3. You will not reach a terminal velocity(follows from 2)
4. The wind will instantaneously accelerate you to 5km/h, and you will maintain that until you hit the ground

It's a pretty easy problem (these assumptions are usual of what you would find in a high school physics class)
First you want to calculate how long it takes you to reach the ground. A trick here is that the wind doesn't affect your vertical velocity, so you can basically ignore it. It will take you the same amount of time to hit the ground if there is wind or not. So to calculate this time you'll use : d = u * t + 1/2 * a * t^2, where:
d = distance travelled
u = initial velocity
t = time
a = acceleration

Remember we are only doing vertical velocity here, so:
u = 0
d = 100 000(we will always work in meters here)
t = ?
a = 9.8

Plug it in and solve and you get t= 142.857 seconds (Sound crazy, but you'll see in a second why it makes sense). Now there are two things left to find, our final vertical velocity and how far horizontally we've travelled. Horizontal distance is easy, we just use distance = velocity * time. Our horizontal velocity is 5km/h, which is 1.38m/s, so distance = 1.38 * 142.857 = 198.41m. We now know how far we've travelled. Finally, the final velocity. So far our velocity is in two components, horizontal and vertical. We know our final horizontal velocity(1.38m/s), now we just have to figure out the final vertical velocity and combine them. Final velocity and be found with: v = u + a * t, where v is the final velocity. Plug everything in:

v = (0) + (9.8 * 142,857)
v = 1400m/s

Because we don't reach a terminal velocity, we keep accelerating at 9.8m/s^2, and when we hit the ground we are traveling 1400m/s vertically! That's 5040km/h, so a time of ~142s doesn't seem so crazy anymore. Anyway now we want our final velocity. We just combine the two in a right angle triangle, calculate the length of the hypotenuse and we have the final velocity (You can find the angle pretty easily as well, I'm just lazy). So:
h^2 = (1.38)^2 + (1400)^2
h = 1400.000689m/s

So to recap we hit the ground 198.41m to the east of where we jumped (100km down) at a velocity of 1400.000689m/s (or 5040km/h). So that's how you would solve it in an ideal physical world. Now all of this is really moot though, because at 100km up you're pretty much in space, and so earth gravity wouldn't be strong enough to pull you down. And let's again say we cut the mountain down to 10km, in real life you would reach a terminal velocity (around ~80km/h I think) and maintain that velocity until you hit the ground, greatly increasing the time. So in that case all you would do is find a humans terminal velocity, find how long it takes to reach that, and repeat the steps above with your new vertical velocity data.

[Tony]

The Earth gravity will very much have an effect on you, even at 100km, just not to as much as it would at the surface. Since the force of gravity is inversely proportional to the square of the distance, then

Earth's radius: ~6,371km
Height: 6,471km (100km from the surface)
Distance increase: 6471/6371 = 1.01569
Gravity decrease: 1.01569^2 = 1.03162618
Gravity at 100km = 9.81 / 1.03162618 ~= 9.5 m/s^2

The way microgravity ("space" orbit) works is that there is enough velocity parallel to the Earth's surface, that you move far enough along Earth's curvature to give yourself more room to continue falling.

"The trick to flying is that you throw yourself at the ground, and miss" -- HHGTTG

Which is exactly what happens, it's a perpetual fall. But since the entire space station is falling at it's terminal velocity, one gets an illusion of 0-gravity (since there is no air friction inside the station).

[kyleisgreat]

You're right, obviously you'll still fall from that height, that was a rather large error on my part.

[ecookman]

o.k assuming that gravity is constant at 100km up (ignoring space)

lol