How Does Physics Explain a Paratrooper's Survival After a 370-Meter Fall?

[uchicago2012]

Homework Statement

In 1955 a paratrooper fell 370 meters after jumping from a plane without his parachute opening. He landed in a snowbank, creating a crater 1.1 meters deep, but suffered only minor injuries. Assuming his mass was 80 kg and his terminal velocity was 50 m/s, calculate
(a) the work done by the snow on him.
(b) the work done on him by gravity.
(c) The work done on him by air resistance as he fell.

Homework Equations

W_non = Change in KE + Change in PE
W_g = m * g * d * cos theta
theta = angle between W_g and the displacement, d

The Attempt at a Solution

for a, I was just wondering if this looked correct:
I thought the force snow exerts on the paratrooper might be nonconservative, like air resistance, since it's sort've like friction. I don't know if that was a good idea.
so then W_non = KE₂ - KE₁ + PE₂ - PE₁
where the initial is at y = 0 (the point at which the guy hits the snowbank - not the point at which he stops moving 1.1 m below the snowbank)
and the final is at y = -1.1, the point at which the guy stops moving
then I solved and found W_non = -100862.4 J

for c,
I'm not sure what to use as my y component
Air resistance is nonconservative so
W_non = KE₂ - KE₁ + PE₂ - PE₁
where the initial is 370 m in the air and the final is 0 or -1.1
if the final is zero, then it's relatively easy. But if it's -1.1 then I have to add in the force of gravity somehow.

Is it as easy as
W_g + W_a = KE₂ - KE₁ + PE₂ - PE₁
where I keep my initial point at 370 m and my final point at -1.1?
Then the work done by my air resistance winds up being considerably more than the work done by gravity.
Which might makes sense, seeing as he survived the fall with minor injuries

 

[PhanthomJay]

Homework Statement

In 1955 a paratrooper fell 370 meters after jumping from a plane without his parachute opening. He landed in a snowbank, creating a crater 1.1 meters deep, but suffered only minor injuries. Assuming his mass was 80 kg and his terminal velocity was 50 m/s, calculate
(a) the work done by the snow on him.
(b) the work done on him by gravity.
(c) The work done on him by air resistance as he fell.

Homework Equations

W_non = Change in KE + Change in PE
W_g = m * g * d * cos theta
theta = angle between W_g and the displacement, d

The Attempt at a Solution

for a, I was just wondering if this looked correct:
I thought the force snow exerts on the paratrooper might be nonconservative, like air resistance, since it's sort've like friction. I don't know if that was a good idea.

That was a very good idea. In Intro Physics (Mechanics), most forces are non conservative, except gravity and spring forces, in particular, which are conservative forces.

so then W_non = KE₂ - KE₁ + PE₂ - PE₁
where the initial is at y = 0 (the point at which the guy hits the snowbank - not the point at which he stops moving 1.1 m below the snowbank)
and the final is at y = -1.1, the point at which the guy stops moving
then I solved and found W_non = -100862.4 J

excellent.

for c,
I'm not sure what to use as my y component
Air resistance is nonconservative so
W_non = KE₂ - KE₁ + PE₂ - PE₁

yes

where the initial is 370 m in the air and the final is 0 or -1.1
if the final is zero, then it's relatively easy. But if it's -1.1 then I have to add in the force of gravity somehow.

There is no air resistance as such in the 1.1 m snowbank crater. That's the snow resistance force , which can be calculated from part a solution.

Is it as easy as
W_g + W_a = KE₂ - KE₁ + PE₂ - PE₁
where I keep my initial point at 370 m and my final point at -1.1?
Then the work done by my air resistance winds up being considerably more than the work done by gravity.

No that equation is incorrect...in several respects...try it using the correct equation, and using the final point as the point where the paratrooper just hits the snow.