How Deep Must Snow Be to Safely Stop a Falling Paratrooper?

  • Thread starter Thread starter patelkey
  • Start date Start date
  • Tags Tags
    Collision Impulse
patelkey
Messages
16
Reaction score
0

Homework Statement



A paratrooper fell 375 m from an airplane without being able to open his chute. Assume that his speed at impact was 52 m/s (terminal speed), that his mass (including gear) was 85 kg, and that the force on him from the snow was at the survivable limit of 1.2 multiplied by 105 N.

a.) What is the minimum depth of snow that would have stopped him safely?

b.)What is the magnitude of the impulse on him from the snow?

Homework Equations



impulse is integration of force with respect to time

The Attempt at a Solution

 
Physics news on Phys.org
patelkey said:

Homework Statement



A paratrooper fell 375 m from an airplane without being able to open his chute. Assume that his speed at impact was 52 m/s (terminal speed), that his mass (including gear) was 85 kg, and that the force on him from the snow was at the survivable limit of 1.2 multiplied by 105 N.

a.) What is the minimum depth of snow that would have stopped him safely?

b.)What is the magnitude of the impulse on him from the snow?

Homework Equations



impulse is integration of force with respect to time

The Attempt at a Solution



The first can be thought of as a conservation of energy problem. Consider how much work it would take to stop the person. (this assumes the distance he falls through the snow adds a negligible amount of potential energy... even if it doesn't you can still solve this in a similar way).

For part b, consider that impulse and momentum have the same units... how much impulse is required to stop X amount of momentum?
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top