A flea jumps 140times their height, what is their height?

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Homework Help Overview

The discussion revolves around a physics problem involving the jumping ability of fleas, specifically focusing on calculating the flea's height based on its jumping distance and take-off speed. The subject area includes kinematics and projectile motion.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore how to determine the flea's height by first calculating the maximum height it can reach based on its take-off speed. There are discussions about using various kinematic equations, and some participants question the assumptions regarding the jump's trajectory and the absence of an angle.

Discussion Status

Some participants have provided insights into the equations relevant to the problem, while others express uncertainty about the calculations and the interpretation of the results. There is an ongoing exploration of the relationship between the take-off speed and the maximum height achieved by the flea.

Contextual Notes

Participants note the lack of an angle for the jump, which complicates the use of the range equation. There is also mention of potential errors in the calculated take-off speed affecting the subsequent height calculation.

SilentBlade91
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Homework Statement


Part 1. Fleas can jump 140times their own height. They jump so they land a max distance of 20.80cm away. What is their take off speed?

I got this part right already, got 1.43m/s what I really need help with is part 2...

Part 2. What is the flea's height?


Homework Equations



I think y=V0y*t-(1/2)gt^2 might be important but I really don't know

The Attempt at a Solution



I'm clueless. I think I have to find out how high the flea jumps and then divide by 140 but I am stumped. Any help is very appreciated!
 
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If the flea jumped straight upward with the take-off speed you calculated, how high would it go?
 
Okay so to find the height the flea jumps, I tried using the formula Vf^2=V0^2+2ad using Vf=0 because it would be 0 at the max height jumped. V0=1.43m/s and a would be gravity but would be negative i believe(-9.8m/s^2). So using that I came up with the height jumped (d) being .104332meters.

So I divided that by 140 and got .000745meters. I'm not sure that is right though I had entered it in the homework and its said it was wrong.
 
Weird, your answer looks okay to me.
 
SilentBlade91 said:

Homework Statement


Part 1. Fleas can jump 140times their own height. They jump so they land a max distance of 20.80cm away. What is their take off speed?

I got this part right already, got 1.43m/s what I really need help with is part 2...

Part 2. What is the flea's height?


Homework Equations



I think y=V0y*t-(1/2)gt^2 might be important but I really don't know

The Attempt at a Solution



I'm clueless. I think I have to find out how high the flea jumps and then divide by 140 but I am stumped. Any help is very appreciated!



Can anyone explain a formula or get me started in how he found the 1.43m/s for the take off speed...I'm practicing questions and I cannot for the life of me get this answer. Please Help!
 
Welcome to Physics Forums.

There is an equation that gives the range (horizontal distance traveled) in terms of take-off speed and launch angle. It should be in your textbook or lecture notes -- that's the easiest way to get the speed.
 
Redbelly98 said:
Welcome to Physics Forums.

There is an equation that gives the range (horizontal distance traveled) in terms of take-off speed and launch angle. It should be in your textbook or lecture notes -- that's the easiest way to get the speed.

thank you...and correct ik the range equation is R= (initial velocity x sin 2theta) / gravity...
where:
R=20.80 cm
initial velocity is what I was looking for
gravity is -9.8 m/s^2
but there was no angle given...unless I am missing something?
 
*I'm just new here*

Looking for some practice problems, I ended up here. ^^

Now, looking into the solutions, I wonder how did you get the take-off speed. Since the equation used for the second question is the right one, the error might come form the take-off speed you've got.
 
Welcome to Physics Forums Meric.
JAK3CAL said:
thank you...and correct ik the range equation is R= (initial velocity x sin 2theta) / gravity...
where:
R=20.80 cm
initial velocity is what I was looking for
gravity is -9.8 m/s^2
but there was no angle given...unless I am missing something?
The 20.80 cm is the maximum range. The largest that the range could possibly be.

To make the range as large as possible, we need sin(2θ) to be as large as possible. What is the largest value that sin(2θ) could have?
 

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