Simple kinematic equations question?

Click For Summary

Homework Help Overview

The discussion revolves around a kinematics problem involving a pole-vaulter falling from a height and landing on a foam-rubber pad. Participants are tasked with calculating the athlete's velocity just before impact and the force exerted during the collision.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants explore the use of kinematic equations to solve for velocity and force, questioning the relevance of the pole-vaulter's mass and the interpretation of the time given in the problem.

Discussion Status

Some participants have provided clarifications regarding the interpretation of the time variable, noting that it pertains to the deceleration after landing rather than the fall duration. There is acknowledgment of misinterpretations, and some have successfully recalculated their answers based on this clarification.

Contextual Notes

Participants express confusion about the separation of the two phases of motion: the fall and the deceleration upon landing. The original poster's attempts indicate a misunderstanding of the problem setup and the application of the kinematic equations.

sbacker
Messages
2
Reaction score
0

Homework Statement



A 52.2 kg pole-vaulter falls from rest from a height of 3.6 m onto a foam-rubber pad. The pole-vaulter comes to rest 0.34 s after landing on the pad.

(a) Calculate the athlete's velocity just before reaching the pad.

(b) Calculate the constant force exerted on the pole-vaulter due to the collision.

Homework Equations



Kinematic equations: V = Vo + AT, X = ((V+Vo)/2)*T, V^2 = Vo^2 + 2AX, X = volt +.5AT^2

F = ma (though I haven't attempted part b yet)

The Attempt at a Solution



It seems like this problem should be easily and quickly solvable using any of the kinematic equations (4 of three values are known, right? Time, distance, acceleration, and original velocity), but this is what I tried and both answers were wrong:

0 = Vo + 9.8 m/s^2 * .34 s (tried 3.33, didn't work)

3.6 = Vo(.34) + .5 * 9.8 *.34^2
3.6 = Vo * .19259 (tried 18.7, didn't work)

I'm a little confused about whether the pole vaulter's mass is irrelevant information, or how to use it—I think I'm a little off with the kinematic equations. Thanks for your help and answers, I know I'm not very good at physics! :)
 
Physics news on Phys.org
Hi sbacker,

sbacker said:

Homework Statement



A 52.2 kg pole-vaulter falls from rest from a height of 3.6 m onto a foam-rubber pad. The pole-vaulter comes to rest 0.34 s after landing on the pad.

(a) Calculate the athlete's velocity just before reaching the pad.

(b) Calculate the constant force exerted on the pole-vaulter due to the collision.

Homework Equations



Kinematic equations: V = Vo + AT, X = ((V+Vo)/2)*T, V^2 = Vo^2 + 2AX, X = volt +.5AT^2

F = ma (though I haven't attempted part b yet)

The Attempt at a Solution



It seems like this problem should be easily and quickly solvable using any of the kinematic equations (4 of three values are known, right? Time, distance, acceleration, and original velocity)

No, for part a you do not know the time. There are two parts to the motion: the fall from the highest point to the pad, and then the motion as the person pushes down into the pad (stopping).

The 0.34seconds refers to how long it takes for the pad to stop the person, not how long it takes for the person to reach the pad.
 
Looks like you've just misread the question. 0.34 seconds isn't how long it takes the pole vaulter to fall onto the pad, it's how long it takes the pad to "pop back" after the vaulter lands. So your first step should be to figure out how long it takes him to reach the pad. The 0.34 seconds don't matter until part b of the question.

Edit: Oops. Thought I'd refreshed the thread before replying.
 
Oh, I read the problem way too fast! How silly of me— thanks for clarifying, I got both parts of the problem right using the kinematic equations with only original velocity, acceleration, and distance. Thanks for pointing out my mistake!
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
8K
Replies
5
Views
3K
Replies
1
Views
2K
Replies
5
Views
4K
  • · Replies 40 ·
2
Replies
40
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K