Free fall and pressure when landing in snow

Click For Summary
SUMMARY

The discussion centers on the physics of parachuting soldiers during WWII, specifically analyzing the safety of dropping soldiers from 150 ft using a dummy bale. The calculations reveal that the pressure experienced upon landing can reach 30 lb/in², which is at the threshold of survivability. The correct approach involves using the equations for work and pressure, with the key formula being F = mgh/Δx. The participants clarify the correct method for calculating time of impact, emphasizing the need for constant acceleration assumptions.

PREREQUISITES
  • Understanding of basic physics concepts, including force and pressure.
  • Familiarity with kinematic equations, particularly those involving acceleration and velocity.
  • Knowledge of work-energy principles in physics.
  • Ability to perform calculations involving units of pressure (lb/in²).
NEXT STEPS
  • Research the implications of pressure on human bodies during high-impact landings.
  • Learn about the physics of free fall and terminal velocity in parachuting scenarios.
  • Explore advanced kinematic equations and their applications in real-world physics problems.
  • Investigate historical military parachuting techniques and their evolution over time.
USEFUL FOR

Physics students, military historians, and safety engineers interested in the dynamics of parachuting and impact forces.

Order
Messages
96
Reaction score
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:
W=F\Delta x
The other one lead to another answer and that is
v=\sqrt{2gh}

The Attempt at a Solution



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

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

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

Now this is two answers and only one can be correct. What is wrong?
 
Physics news on Phys.org
Hi Order! :smile:

Your equation ∆t = ∆x/v is wrong.

∆t = ∆v/a would be correct (if we assume constant acceleration ). :wink:
 
tiny-tim said:
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 \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} which leads to \Delta t = \frac{2\Delta x}{v} Now this must be correct. Thanks for help!

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

Similar threads

  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 30 ·
2
Replies
30
Views
4K
  • · Replies 21 ·
Replies
21
Views
5K
Replies
12
Views
6K
Replies
15
Views
4K
Replies
17
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
5K
Replies
6
Views
4K