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Free fall and pressure when landing in snow

  • Thread starter Order
  • Start date
  • #1
97
3

Homework Statement



During the second world war the Russians, lacking sufficient parachutes for airborne operations, occasionally dropped soldiers inside bales of hay onto snow. The human body can survive an average pressure on impact of 30 lb/in2.
Suppose that the lead plane plane drops a dummy bale equal in weight to a loaded one from an altitude of 150 ft, and that the pilot observes that it sinks about 2 ft into the snow. If the weight of an average soldier is 144 lb and his effective area is 5 ft2, is it safe to drop the men?

Homework Equations



I use two equations. One lead to the correct answer and that is the formula:
[tex]W=F\Delta x[/tex]
The other one lead to another answer and that is
[tex]v=\sqrt{2gh}[/tex]

The Attempt at a Solution



I use the notation [tex]h=150 ft[/tex] [tex]m=144 lb[/tex] [tex]\Delta x = 2 ft[/tex] [tex]A=5ft^{2}[/tex]

1. The work done on the falling body is zero from jump to landing so [tex]0 = mgh-F\Delta x[/tex] and therefore [tex]F=\frac{mgh}{\Delta x}[/tex] The pressure is therefore [tex]P=\frac{F}{A}=\frac{mgh}{\Delta xA}= 15 lb/in^{2}[/tex]

2. The speed of the falling body at impact is [tex]v=\sqrt{2gh}[/tex] As it hits the snow it will drop in speed during a time [tex]\Delta t = \frac{\Delta x}{v}[/tex] Therefore the force on the body at landing will be [tex]F=\frac{mv}{\Delta t}[/tex] This leads to a pressure of [tex]P=\frac{mv}{\Delta t A}=\frac{2mgh}{\Delta x A} = 30 lb/in^{2}[/tex] Very risky indeed.

Now this is two answers and only one can be correct. What is wrong?
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
25,832
251
Hi Order! :smile:

Your equation ∆t = ∆x/v is wrong.

∆t = ∆v/a would be correct (if we assume constant acceleration ). :wink:
 
  • #3
97
3
Hi Order! :smile:

Your equation ∆t = ∆x/v is wrong.

∆t = ∆v/a would be correct (if we assume constant acceleration ). :wink:
Yes, I think we assume constant acceleration since we assume constant pressure on the body. Ok, so a=-d2x/dt2 leads to [tex]\Delta x=v \Delta t -\frac{a\Delta t^{2}}{2}=v \Delta t -\frac{v\Delta t^{2}}{\Delta t2}=\frac{v \Delta t}{2}[/tex] which leads to [tex]\Delta t = \frac{2\Delta x}{v}[/tex] Now this must be correct. Thanks for help!

I thought there was something wrong and thought about average velocities , but this was much better.
 

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