OpenStudy (mendicant_bias):

I'm just not sure which approach I should take with this. I've tried applying a few equations (below) and none of them (from what i'm seeing) have the information needed to solve for a single variable. "A frisbee is lodged in a tree branch 6.2 meters above the ground. A rock thrown from below must be going at least 3.0 m/s to dislodge the frisbee. How fast must such a rock be thrown upward if it leaves the thrower's hand 1.1 meters above the ground?"

5 years ago
OpenStudy (mendicant_bias):

Equations:$v = v _{o}+at$$x=x _{o}+vt+\frac{ 1 }{ 2 }at ^{2}$$v ^{2}=v _{o} ^{2}+2ax$

5 years ago
OpenStudy (jamesj):

So the rock travels a distance of x = 6.2 - 1.1 = 5.1 m before hitting the frisbee. You know when the does hit the frisbee it must be travelling at v = 3.0 m/s You want to find v_0

5 years ago
OpenStudy (jamesj):

You need to fill in the blank with one more fact and then you can use one of these equations.

5 years ago
OpenStudy (jamesj):

Following?

5 years ago
OpenStudy (mendicant_bias):

Yes, sorry, wasn't on earlier. I'll look at what you said, but I was just checking in this minute to see if I got a reply. Thanks!

5 years ago
OpenStudy (mendicant_bias):

Or I might have enough time now? Time will tell, no pun intended. I did the same thing you just did (equate the frisbee with 5.1 m) and from there can assume x (naught) = 0. From there, I can find that:$5.1 = 0 + vt +\frac{ 1 }{ 2 }at ^{2}$aaand i've got to go right now. I think my actual problem is the way i'm interpreting how it was written, e.g. "must be going at least 3 m/s". bbl.

5 years ago
OpenStudy (jamesj):

I have give you x and v You also know a. (Remember gravity?) And you want to find v_0 Which equation has x, a, v and v_0? Use it!

5 years ago
OpenStudy (mendicant_bias):

Yeah, it was just an issue with the wording.$v ^{2}=v _{o}^{2} + 2ax$Given:$9=v _{o}^{2}+2(-9.8)(5.1)$Wait, what? Can't get a nonreal answer (negative square root). Is this just a reference frame problem, i.e. should a really just be |g|?

5 years ago
OpenStudy (jamesj):

This doesn't imply $$v_0^2$$ is negative ...

5 years ago
OpenStudy (mendicant_bias):

I'm aware of that, but you can't take the square root of a negative number...? I understand what's implied. That doesn't mean that I can't question why it doesn't mesh with the math.

5 years ago
OpenStudy (mendicant_bias):

(Or, you can't get a real answer with a negative square root, which this definitely has to be.)

5 years ago