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

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The discussion revolves around calculating a flea's jump height using basic physics principles. The initial problem presents a flea reaching a takeoff speed of 1.0 m/s over a distance of 0.50 mm, leading to questions about its acceleration, duration of the acceleration phase, and maximum height. The book's answers include an acceleration of 1000 m/s², a time of 0.0010 s, and a height of 5.1 cm, which some participants believe contain errors. Participants clarify that gravity must be considered after the flea's initial acceleration phase, emphasizing that the calculations should reflect the effects of gravity on the flea's jump. Accurate application of physics equations is essential for determining the correct jump height.
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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?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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