Impulse Momentum Theorem for Calculating Fall Height

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SUMMARY

The discussion focuses on applying the Impulse Momentum Theorem to calculate the fall height of a student weighing 63 kg who falls freely and comes to rest after a collision with the ground. The average force exerted by the ground is +16000 N, and the time of impact is 0.018 seconds. The correct approach involves determining the final velocity using the impulse-momentum relationship and then applying kinematic equations to find the height from which the student fell, ultimately leading to a calculated height of approximately 1.066 meters.

PREREQUISITES
  • Understanding of the Impulse Momentum Theorem
  • Basic knowledge of kinematic equations
  • Familiarity with gravitational acceleration (9.8 m/s²)
  • Ability to perform calculations involving forces and motion
NEXT STEPS
  • Review kinematic equations for uniformly accelerated motion
  • Study energy conservation methods in mechanics
  • Learn how to apply the Impulse Momentum Theorem in various scenarios
  • Explore the relationship between force, mass, and acceleration using F=ma
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators looking for practical examples of the Impulse Momentum Theorem and kinematic calculations.

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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