Recent content by AC5FFw

Lumen... Your graph/pic did not come up.. but that's more of a issue with my server here.. All sorts of restrictions. I will check this out when I get home this evening... Thank you all! I'm definatly a lot smarter on this now than I was yesterday! LOL Where were all these great...

I'm on board now! :D I also just learned about the quadratic equation as well. I definatly KNOW I was figuring that equation wrong now. After learning that bit on the quadratic and working the numbers through it I get the same 10.4 result. I'm a happy camper now! :)

Plugged this formula into the quadratic equation and got my 10.4 as one of the results. With the +/- option, I take it I have to figure out both then plug them back into the equation. One works and the other does not. So I know that "t" has to equal 10.4 Sound about right?

Just reading on the quadratic equation now. I don't think I was ever taught this. So, when I look at an equation like the one above I try to see how to simplify it further. WoW what Fun!

Don't think so. Guess I could google it.

While a problem, this is more for me to learn to help my kid with her own homework. I used to be great at this sort of thing but for some reason I just can't get my head around this one. -125 = 39.1t - 4.9t^2 I need to find the value for t. It should be approx 10.4. I would just like to...

I am SO glad you two are not here with me. You would probably slap me silly. LoL RL: Your analogy makes perfect sense. I understand your zero displacement. So in this case, because you are basically startng at the same "Y" position on the graph, the Voy would be the same as when the object...

Wouldn't that just be the runner's acceleration divided by the cheetah's?

I think I'm confused more now! :D I talked w/my daughter tonight, and the correct answer for total flight time is 10.4. Anden you are right.. But this is what confuses me. I will try to find the website I was reading on how to figure this, but from my knowledge, I would think that Voy would...

Anden.. Okay - I'm with you now. However, let's go back to your original solution; -125 = Voy t - 0,5gt^2 gives t = 10,4 But did you miss - or am I missing the extra 78m it rose before falling in your equation? Should not the -125 be instead 203. If so, then my 14.4s is correct. I...

okay. are you using 9.8m/s/s for g or are you using -9.8m/s/s? Maybe that is the difference? The papers that I have been reading show to use a -9.8. Back to my original problem though; looking at result "C" the vertical and horizontal velocity at impact. The vertical is going to be...

Something is different here... You have Y=(Voy)(t)-(0.5)(g)(t^2) Everything I ahve seen so far is: Y=(Voy)(t)+(0.5)(g)(t^2) What is the difference between these two? I have not seen this formula expressed with the - sign

Wouldn't (d) be positive? That's how I got the fall time of 14.4. But, if I take this as a -125, then add the rise of 78 and (1/2a), gives 51.8 then the fall time is 7.2s This is a lot closer to what I would think it should be.. HOWEVER: if my time to peak of 3.98s is right, that means...

not sure of my equations here, so... If I am right and the max hight is 203m (inital 125m + y-displacement of 78m)then: 203=(.5)(a)(t^2) 203=(.5)(-9.8)(t^2) 207.9=(t^2) 14.4s = t (approximate) So, overall flight time = 18.38sec I'm fairly sure of the time to peak of trajectory...

I think I'm on the right track.. Thx for the help! Just to verify, here's what I got so far. Vox=51.9m/s Voy=39.1m/s t (peak of trajectory)=3.98s dy (y-displacement)=78m From this I know that it travels 7.96s until it is at the same hight as it left. I need to figure the rest since...