Calculating the Height of a Flea's Jump Using Basic Physics Principles

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Homework Help Overview

The discussion revolves around calculating the height of a flea's jump using basic physics principles, specifically focusing on acceleration, time duration of the jump, and the maximum height achieved. The problem involves kinematics and the effects of gravity on the flea's motion.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the calculation of acceleration and its implications for the jump height. There are questions regarding the treatment of gravity in the calculations and whether the provided acceleration value should be used in conjunction with initial velocity to determine the jump height.

Discussion Status

Participants are exploring different interpretations of the problem, particularly the role of gravity in the calculations. Some have provided calculations based on the acceleration derived from the problem statement, while others are questioning the assumptions made about gravity and its impact on the flea's jump.

Contextual Notes

There is mention of potential typos in the book's answers, particularly regarding the units of acceleration and the height calculation. Participants are also considering the implications of using the calculated acceleration versus the acceleration due to gravity in their reasoning.

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Homework Statement


I did this problem in my book, then checked the answer, and it showed what must be a typo for answer a (acceleration should be m/s^2 not m/s), and answer c just makes no sense.

"When jumping, a flea reaches a takeoff speed of 1.0m/s over a distance of 0.50mm."

a. What is the flea's acceleration during the jump phase?

b. How long will does the acceleration phase last?

c. If the flea jumps straight up, how high will it go?

The book's answer's
a. 1000m/s
b. 0.0010s
c. 5.1cm

Homework Equations



a.)

a= (((vf)^2-(vi)^2) / (xf-xi)) / 2

a= (((1.0m/s)^2-(om/s)^2) / (5.0e-4m - 0m)) / 2

b.)

(Change in time)= (vf -vi) / a

c.)

Yf= Yi +vi(change in time) + 1/2 (a)(change in time)^2

The Attempt at a Solution


a.
a= (((1.0m/s)^2-(om/s)^2) / (5.0e-4m - 0m)) / 2
= 1000m/s^2

b.
(Change in time)= (1.0m/s - 0m/s) / 1000m/s^2
= 0.001s

c.
Yf=0m +0m/s(0.001s-0s) + 1/2 (1000m/s^2)(0.001s-0s)^2
=5.0e-4m
 
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Bubofthedead said:

Homework Statement


I did this problem in my book, then checked the answer, and it showed what must be a typo for answer a (acceleration should be m/s^2 not m/s), and answer c just makes no sense.

"When jumping, a flea reaches a takeoff speed of 1.0m/s over a distance of 0.50mm."

a. What is the flea's acceleration during the jump phase?

b. How long will does the acceleration phase last?

c. If the flea jumps straight up, how high will it go?

The book's answer's
a. 1000m/s
b. 0.0010s
c. 5.1cm


Yf=0m +0m/s(0.001s-0s) + 1/2 (1000m/s^2)(0.001s-0s)^2
=5.0e-4m

In part c how did you account for gravity?
 
I used the 1000m/s^2 as acceleration instead of gravity, unless the 1000m/s^2 was meant to be used to somehow solve for vi.
 
Bubofthedead said:
I used the 1000m/s^2 as acceleration instead of gravity, unless the 1000m/s^2 was meant to be used to somehow solve for vi.

I'm sorry but you can't change gravity.

Your acceleration was just to get to speed over the tiny distance of the fleas' little legs. After that the flea is in the hands of gravity.

If your initial velocity is 1 m/s how high will it go?

Any equations that pop to mind?
 

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