Solve 1-D Kinematic Flea Jump Problem

[QuarkCharmer]

Homework Statement

If a flea can jump straight up to a height of 0.510 m, what is its initial speed as it leaves the ground? How long is it in the air?

Homework Equations

The four kinematics equations with constant acceleration I think.

The Attempt at a Solution

I let the positive y-axis be the jump height and the x-axis time.

From there I list some known variables:
[itex]a_{y} = -9.8 m/s^{2}[/itex]
[itex]y_{0} = 0 m[/itex]
[itex]\Delta y = 0.510 m[/itex]

I'm not really sure how to get it from here. I am assuming that he reaches the apex of y=0.510 meters at 1/2 the total air time. But I can't figure out which equation to use in this case.

 

[QuarkCharmer]

I am pretty sure this is the equation to use here.

[tex]v_{f}^{2} = v_{0}^{2} + 2(a)d[/tex]

Do I assume his final velocity is at the apex of the jump, thus?

[tex]0^{2} = v_{0}^{2}+ 2(-9.8)(0.510)[/tex]

So his initial velocity would be:

[tex]v_{0} = \sqrt{2(9.8)(.510)}[/tex]

 

[lewando]

Since you don't know time, which of the 4 equations does not rely on time?

[edit] You beat me to it. Looks like you are on the right track!

 

[QuarkCharmer]

Since you don't know time, which of the 4 equations does not rely on time?

The one above, making initial velocity = 3.16 (to two significant figures)I think that opens up the possibility of using this one:

[tex]x(t) = x_{0} + tv_{0} + \frac{at^{2}}{2}[/tex]
?

So:
[tex].510 = 0 + 3.16t + 4.9t^{2}[/tex]

therefor t= .13368 (about 0.134 seconds)

That seems like a small timeframe to me, which is why I am wondering if I am on the right track here.

Edit: Yeah that is incorrect, I don't see what I am doing wrong here. Oh wait it should be -9.8, giving it another shot..

 

[QuarkCharmer]

Okay, that can't be it because that quadratic equation would not be solvable.

 

[QuarkCharmer]

Oh my, I had the right time, I just needed to multiply it by 2 to represent the other 1/2 of the jump because t was giving me the time at the top...

Using -9.8 in the above gave me t = .3224 so really it was 6.448

Ugh.

Thanks

 

[lewando]

You need to be careful when evaluating -4.9t² +3.16t -0.51 = 0. The B coefficient in not "3.16", but rather [tex]\sqrt{2*9.8*0.51}[/tex] That's why the imaginaries pop up with calculators, applets, etc.

Keep it simple: v = v₀ + at gets you there also.

Using -9.8 in the above gave me t = .3224 so really it was 6.448

I think you meant round trip time is 0.6448 seconds