Solve Impulse and Depth Homework Challenge

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The discussion revolves around a physics homework problem involving a paratrooper who fell 370 meters and landed in snow, requiring calculations for the minimum snow depth to stop him safely and the impulse from the snow. Participants emphasize the importance of using relevant equations, such as impulse and acceleration formulas, to solve the problem. There is some tension as users express frustration over the original poster's urgent tone and lack of initial contribution to the discussion. Key equations mentioned include F=ma and impulse calculations, with users helping clarify the relationship between force, mass, and acceleration. The conversation concludes with users exchanging gratitude for assistance and discussing their academic backgrounds.
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[SOLVED] Impulse and depth********!

Homework Statement




In February 1955, a paratrooper fell 370 m from an airplane without being able to open his chute but happened to land in snow, suffering only minor injuries. Assume that his speed at impact was 58 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 1.2 105 N.

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

?m

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


? kg·m/s


Homework Equations




I have no clue how to do this problem

The Attempt at a Solution



I DONT KNOW. i am in desparate need of help.


ANY CLUES or ANY hints will help.

THANK YOU!*****************
 
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this is URGENT, I CAN USE ANYONE'S HELP IF THEY ARE WILLING TO LEND ME A HAND FOR THIS PROBLEM, PLEASE!
 
In February 1955, a paratrooper fell 370 m from an airplane without being able to open his chute but happened to land in snow, suffering only minor injuries. Assume that his speed at impact was 58 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 1.2 105 N.

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

?m

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


? kg·m/s

PLEASE HELP ON THIS PROBLEM
ANY HELP ON THIS PROBLEM WILL BE GREATLY APPRECIATED
 
Ok, so what's big panic? It's just a physics question. You have read the forum rules, right? You have to contribute first. For example, how is an initial velocity and a final velocity and a stopping distance related to acceleration? You can find a formula someplace I'm sure. Ordinarily, I would have stopped at the fourth sentence, so I'm giving you extra patience. Don't push me. We don't get paid for doing this.
 
Dick said:
Ok, so what's big panic? It's just a physics question. You have read the forum rules, right? You have to contribute first. For example, how is an initial velocity and a final velocity and a stopping distance related to acceleration? You can find a formula someplace I'm sure. Ordinarily, I would have stopped at the fourth sentence, so I'm giving you extra patience. Don't push me. We don't get paid for doing this.

first off, I am stressed because this problem is worth a lot to my grade--you should not worry about that. and don't degrade me or talk down to me.
um for your info. i have contributed. if you don't want to help me then don't, but you do not have to get mad. i don't need your patience and what do you mean by "don't push me" stop being such a jerk, I am not stupid, i know that no one get paid.
 
Fine, I understand you are stressed. Sorry. I'm not mad, you apologized. Now I'm just trying to help. But you haven't contributed yet, you have just posed a question without telling us what you know. Ok, I'm telling you that there is an equation that relates an initial velocity and a final velocity and a distance in between with the acceleration. Can you find it? It involves velocities squared and an acceleration times a distance. Calm down.
 
basically, he wants us to do it..
 
yes, i know that equation but see for the problem you don't need to find out acceleration.
anyways, this is what you want me to do: v2-v = 2as so acceleration = 4.54 but you see the question is not asking for the acceleration
 
so now what do i do?
i know that impulse = Force multiplied by time but i know neither
and also we have to find out the minimum depth. i don't know how to do eith erone
 
  • #10
can either of you help me now? :(
 
  • #11
Acceleration has a close relation with force. F=ma.
 
  • #12
Im in ap physics C this year, lol i know ur pain.. but see.. life isn't fair... I joined this forum mainly cause i just love physics and i like helping ppl out, and learn from the experience. I mean, I'm a student too.
Don't expect to just join the forum and paste homework or w/e and then write in bold letters that u need immediate help. I'm trying to help u out man. Personally i don't care, but when ppl see that kind of selfishness they will ignore ur threads.. so you keep that in mind..

ok now in that problem...

u know F=ma, u are given a force and mass, find the acceleration.. u know the velocity before the impact and the velocity after it stops (0). Also, you can assume that the person is falling exactly vertically with no velocity in the x component (very poor assumption because planes aren't stationary). That should be enuf.. show some work.
hint: look for an equation which has velocity, acceleration and distance.

if u are still confused, i suggest looking in wikipedia.org instead of yelling at us for not helping
 
  • #13
hey Dick, i might need ur help for the second part. i mean, i know what to do, but see it gets a bit confusing cause i have learned that impulse= mV1-Mv2, but in wiki it says impulse is F(delta T), but i always -F(delta T). huh could u tell me which one it was again? I'm forgetting.
 
  • #14
vi^2-vf^2=2*a*s. Yes. But how can you say a=4.54 in whatever units? You don't know s yet. Compute a from F=ma, where F is the maximum survivable force. Put a and the velocity into that and find s.
 
  • #15
yup, but Dick, did u get my post lol?
 
  • #16
aq1q said:
hey Dick, i might need ur help for the second part. i mean, i know what to do, but see it gets a bit confusing cause i have learned that impulse= mV1-Mv2, but in wiki it says impulse is F(delta T), but i always -F(delta T). huh could u tell me which one it was again? I'm forgetting.

m*v1-m*v2=m*(v1-v2)=m*a*t=F*t. Whether it's +F*t or -F*t depends on whether you are talking about the impulse transmitted from 1 to 2 or the impulse transmitted from 2 to 1. Just like forces. And like I pointed out to the OP it is rude to interject you own problem into somebody elses post.
 
  • #17
lol I am pretty sure... dick and i are saying the same thing.. and he didn't say that a is 4.54 and i never said a is 1411.764
 
  • #18
hey physicsbhelp what is the force again.. i can't really understand is there a decimal after 1?
 
  • #19
ooooo okay i get it now! thakns so that is the answer to the first part of the question right? so s = 1.1914
im sry I am a little slow at physics. please forgive me.
so then how do i find time because i have to ifnd impulse now.
 
  • #20
i don't know cause lol i can't read the force it says "1.2 105N" why is there a space
 
  • #21
physicsbhelp said:
yes, i know that equation but see for the problem you don't need to find out acceleration.
anyways, this is what you want me to do: v2-v = 2as so acceleration = 4.54 but you see the question is not asking for the acceleration

What's with you guys, are you in the same class or something? He did say a=4.54. Look, I got to go.
 
  • #22
um no.. dick.. wtf?
 
  • #23
okay thanks.
and um the force is 1.2 times 10 to the 5th power.
 
  • #24
aq1q said:
i don't know cause lol i can't read the force it says "1.2 105N" why is there a space

The OP meant 1.2*10^5N. I've seen this before, so I just guessed.
 
  • #25
1.2e5
1.2 E 5
 
  • #26
exactly.
 
  • #27
1.2 is what I am getting.. gj
 
  • #28
physicsbhelp said:
okay thanks.
and um the force is 1.2 times 10 to the 5th power.

Ok, and try to stress less, ok? Now I'm all nervous.
 
  • #29
ok.. but what does that have to do with him and i being in the same class?
 
  • #30
am i going to use Sy = VyT + g(t squared) times .5
 
  • #31
i rounded of course.. sigfigs.. 1.1914 = 1.2
 
  • #32
okay thanks that's what i got, but now for teh second part am i going to use Sy = VyT + g(t squared) times .5
 
  • #33
well first of all u can use that.. but u won't use g.. u will use a, which u solved for.. remember 1411.76...however that's the longer way.. do it this way

Vf=Vo+at (sorry if i use a and V, that's the way i am taught, instead of S)
 
  • #34
aq1q said:
ok.. but what does that have to do with him and i being in the same class?

Apologies. I'm not used to getting hammered by two people at the same time. I get confused. You seem to be asking the same questions and I thought you might be working on the same problem. This is a really good reason not the break into somebody else's thread with your own problem.
 
  • #35
the final velocity =0, initial velocity= known, once u get the time.. u know what to do next.
 
  • #36
Bye.
 
  • #37
okay so for time i got = 2.38s is that right>?
 
  • #38
Dick said:
Apologies. I'm not used to getting hammered by two people at the same time. I get confused. You seem to be asking the same questions and I thought you might be working on the same problem. This is a really good reason not the break into somebody else's thread with your own problem.

well yes.. but it wasn't really a problem it was just something i needed reassurance. and today is my first day here.. I'm learning. my apologies
 
  • #39
but i don't know the force so how can i find impulse.
 
  • #40
im not sure, but that doesn't look right.. what did u do?
 
  • #41
force is given
 
  • #42
i did V= Vinitial + at and for V initial i used 58
ooooooo wait i accidentaly squared 58...
so now for t I am getting = .04108s
 
  • #43
there is an easier way to do this.. impulse= f(delta T), but it is also = mV1-mV2
 
  • #44
so is the impulse = 4930.002?
 
  • #45
or do i multiply 4930 by 1.2 E 5?
 
  • #46
ya its 4930
 
  • #47
1.2 times 10 to the 5th power
 
  • #48
how did u get that?
 
  • #49
Omg
Thank You!
You Are The BEST!
!
!
!
!
 
Last edited:
  • #50
Nvm What I Said In Post 64.
 
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