Impulse Momentum Theorem for Calculating Fall Height

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Homework Help Overview

The discussion revolves around applying the impulse momentum theorem to determine the height from which a student fell after striking the ground. The problem involves concepts of forces, acceleration, and kinematics within the context of free fall and collision dynamics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of the impulse momentum theorem and question the inclusion of gravitational force in their calculations. There are attempts to relate force, mass, and acceleration to find the height of the fall, with some participants expressing confusion about the correct approach.

Discussion Status

Participants are actively engaging with the problem, exploring various interpretations and methods. Some have provided corrections and clarifications regarding the forces involved and the nature of the motion during the fall. There is a recognition of the need to differentiate between the impact duration and the fall time.

Contextual Notes

Participants note that the problem states to assume only the force from the ground acts during the collision, which influences their calculations. There is also mention of the acceleration due to gravity and its role in determining the height fallen.

ladolce
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Impulse Momentum Theorem!

Homework Statement


A student (m = 63 kg) falls freely from rest and strikes the ground. During the collision with the ground, he comes to rest in a time of 0.0180 s. The average force exerted on him by the ground is +16000. N, where the upward direction is taken to be the positive direction. From what height did the student fall? Assume that the only force acting on him during the collision is that due to the ground.


Homework Equations



impulse momentum theorem

The Attempt at a Solution



i did it and got vfinal=4.57 and put that over the time, 0.018sec and got 254 m

but that's not right... i don't know why...anyone able to help?
 
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ladolce said:
i did it and got vfinal=4.57
Did you include the force of gravity?
and put that over the time, 0.018sec and got 254 m
No idea what you did here. If something fell from that height, how fast would it be going?
 
I'm a little lost on how to approach this problem :(
 
Once you've found the speed, finding the height is just a falling body problem.
 
Ok, so can I just use F=ma and say 16000 N=63kg x (a) and get acceleration and then use kinematics to say

displacement=1/2 (a) (0.018)sqrd?
 
No. You are finding the height the student fell from to reach the speed that you figured out. (You do not need the displacement during his interaction with the ground.) The acceleration is the acceleration due to gravity.

How did you find the speed?
 
I made 16000=63v/.018 and got 4.57 :/
 
ladolce said:
I made 16000=63v/.018 and got 4.57 :/
That's the impulse momentum theorem--good. But realize that you need the net force, not just the force from the ground. What other force acts on the student? (This correction will give you a slightly different speed.)
 
Ah, weight=mg=617.4

so i get v=4.395
 
  • #10
what could i do next?
 
  • #11
Oh, in the problem it says : "Assume that the only force acting on him during the collision is that due to the ground."

so that's all i have to worry about. i have the right velocity as 4.57 m/s then...i don't know what to do next however.
 
  • #12
Now figure out the height he must have fallen from. This part has nothing to do with the collision with the ground or with the impulse-momentum theorem.
 
  • #13
I keep getting really small answers. Like, wouldn't I just do 4.57m/s=x/.018sec?
 
  • #14
ladolce said:
I keep getting really small answers. Like, wouldn't I just do 4.57m/s=x/.018sec?
This is not correct. The 0.018 sec is the duration of the impact with the ground, not the time it takes him to fall. (Also: Falling is accelerated motion, not constant speed.)

Review your basic kinematic relationships. (Hint: You can also use energy methods, if you like.)
 
  • #15
Ok, I did it again and got

y=(0)-(4.57 squared)/(2)(-9.8)

and i got -1.066m

would that be what you got?
 
  • #16
Wow, that's right. Thank you so much for your time once again!
 
  • #17
Sounds good to me. (Lose that minus sign.)
 

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