Recent content by michelleri

Got the PM's. Will work it all out. Thanks again very much.

AWESOME! Thanks again. Good night.

I am hanging on...

Ok I think I got it. I am not sure if you have been solving it or not. I got 1.52 m. He doesn't put a lot of weight on the actual number, and I have every step detailed. One more question and you can sleep :) Part B, is it the vertical component we already figured out at .497 m/s?

scratch that, the time is .84 s (thanks for the tip on the internet solver!)

ok so it will be: xf-xo = voxt where vox is 1.81 m/s, t is 2.35 s and I am solving for xf? Is xo zero?

We are allowed the calculators, but I don't have one. I will do it the old fashioned way. Once I get t, how is that going to help me figure the distance the trout traveled horizontally?

sent you a pm as well

Equations I have so many equations. Here they are: Vyf2 = voy2 – 2g (yf-yo) Yf – yo = voyt – ½ gt2 Xf – xo = voxt Vyf = voy - gt I typed them up nicelt using word. It is attached.

hold on one sec...

Yes. I that what I was just doing after your first post. I fugured Uv to be .515 m/s and Uh to be 1.95 m/s. Is that right?

No problem! THANKS

I figured them out. The velocity is 1.88 m/s and the distance is 3.89 m

This is exactly what was given to me, word for word. I guess it would be free falling. How can I figure out the x component when I don't know the x distance? This problem is way over my head. This is my first physics class, I think this is a little difficult for Intro to Physics. Can you...

Velocity using quadratic formula- NEED HELP PLEASE! Here is the problem: An osprey is flying over a reservoir with a trout in her talons. Her speed is 4.21 mph and the reservoir is 12.76 feet below. She is flying at an angle of 15.32 degrees below the horizontal- A- Calculate the...