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

  • #1

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!!
 

Answers and Replies

  • #2
Redbelly98
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If the flea jumped straight upward with the take-off speed you calculated, how high would it go?
 
  • #3
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.
 
  • #4
Redbelly98
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Weird, your answer looks okay to me.
 
  • #5
6
0

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!
 
  • #6
Redbelly98
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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.
 
  • #7
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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 im missing something?
 
  • #8
1
0
*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.
 
  • #9
Redbelly98
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Welcome to Physics Forums Meric.
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 im 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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