Free fall and pressure when landing in snow

AI Thread Summary
The discussion centers on the safety of dropping soldiers from a plane at 150 feet onto snow, using a dummy bale as a test. Calculations show that the pressure experienced upon landing can be either 15 lb/in² or 30 lb/in², leading to confusion over which result is accurate. The initial approach incorrectly used the equation ∆t = ∆x/v, which was pointed out as flawed. The correct approach involves assuming constant acceleration, leading to a revised equation for ∆t. Ultimately, the discussion emphasizes the importance of accurate calculations in assessing the risks of such parachute drops.
Order
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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?
 
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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.
 
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