Some simple physics

[Guest]

So lets say Will jumps from an airplane that is at X hight. It took him Y minutes to reach the ground.

Is it safe to assume that if he jumps from 2X hight, that it will take him 2Y minutes to touch ground?

How about 5X and 5Y?

Or 100X and 100Y?

[bonanova]

So lets say Will jumps from an airplane that is at X hight. It took him Y minutes to reach the ground.

Is it safe to assume that if he jumps from 2X hight, that it will take him 2Y minutes to touch ground?

How about 5X and 5Y?

Or 100X and 100Y?

No.

And I hope he brought a parachute.

His speed the first X [feet, let's say] is less than his speed the 2nd X feet,

so it will take him less than 2Y minutes jumping from 2X height.

In general, ignoring wind resistance, Y increases as the square root of X.

For 2X, the time is Y x sqrt[2] = about 1.414Y.

For 5X, the time is Y x sqrt[5] = about 2.236Y.

For 100X, the time is 10Y.

[Guest]

you have the right answer with good reasoning.

But actually, I was lookin' for the part with wind resistance.

At a greater hight, Will will spend more time in terminal velocity,

raising his speed average.

But yeah, I understand what you mean

[Guest]

Well, it depends on how big is X.

His speed the first X [feet, let's say] is less than his speed the 2nd X feet,

so it will take him less than 2Y minutes jumping from 2X height.

In general, ignoring wind resistance, Y increases as the square root of X.

For 2X, the time is Y x sqrt[2] = about 1.414Y.

For 5X, the time is Y x sqrt[5] = about 2.236Y.

For 100X, the time is 10Y.

That is true, but for a fixed gravitational force. So, if X is a very big number it is going to be much slower. For instance if X = 3,844 km then 100X=384,400 km (which is almost the same as the distance of the moon from Earth) sure it is not going to be 10Y b/c from certain onwards the gravitational force gets smaller and smaller.

[bonanova]

Is it safe to assume that if he jumps from 2X hight, that it will take him 2Y minutes to touch ground?

Wind resistance eventually decreases his acceleration to zero. [terminal velocity]

Starting great distances from the earth, gravity is initially weaker.

Both influences taken into account, however, his speed never decreases.

Since he starts with zero [downward] velocity, the first half of his fall will take longer than the second half does.

I think, in all cases, is it will take less than 2Y minutes to fall 2X distance.

Two caveats:

Closer to earth, [1] wind resistance increases, decreasing the terminal velocity, and [2] gravitational pull increases, tending to increase terminal velocity. Because the thickness of the atmospheric layer is a small fraction of the earth's radius, the first effect is stronger.

His terminal velocity I believe actually does decrease.

But not enough to make up for zero initial velocity, I think.