
#1
Sep2305, 09:00 AM

P: 33

I'm having problems with this question. I know it should be simple, but I can't seem to get the right answer.
In this problem, you will use kinematic equations applied to a flea in free fall. Take the magnitude of freefall acceleration to be g = 9.80 m/s^2. If a flea can jump straight up to a height of 0.550 m, what is its initial speed v_0 as it leaves the ground? My answer was 3.28 m/s, which is correct. How long is the flea in the air? I don't know which equations to use. to reach it's max height, I used: t=d/v and got 0.168s to reach the ground, I used: y1 = 0.5gt^2 and got 0.335s so the total time I calculate was 0.503s Can somebody help me? Thanks 



#2
Sep2305, 09:19 AM

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P: 2,538

There are a couple of ways to go about this: For example, you could try using [tex]0=v_0t\frac{1}{2}gt^2[/tex] for the entire jump. P.S. Do you think it's odd that the flea took longer to fall than to rise? 



#3
Sep2305, 09:21 AM

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P: 2,004

You did use the right equation when it was falling down: y=1/2gt^2. If you use symmetry, you know the flea takes as much time going up as it does falling down. 



#4
Sep2305, 09:22 AM

Math
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Thanks
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P: 38,896

How long is the flea in the air for?
How did you get 3.28 m/s as the initial speed? I would assume you had to find the time until the greatest height in order to get that if so, because of the symmetry, you only have to double that to find the total time in the air.
Or you could use h(t)= vt 4.9t^{2}= 0 when the flea is back on the ground, using the value you got for v, and solve for t. Your "t=d/v and got 0.168s" is wrong because that only works for constant velocity. Since the velocity at the top of the jump is 0, "y1 = 0.5gt^2 " does work for the time back to floor. It should be clear that, ignoring air resistance (which for a flea is not very realistic) the time from the floor to the top of the jump is exactly the same as the time from the top of the jump back to the floor. 



#5
Sep2305, 10:10 AM

P: 33

Thank you all very much for the help...and quick replys!
I am glad that I found this board. How did you get 3.28 m/s as the initial speed? I used the equation: v1^2 = vo^2  2g(delta_y) 


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