Velocity using quadratic formula- NEED HELP PLEASE

[michelleri]

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 trout's range, that is, how far does the trout travel horizontally before splashing? Neatly show your work with units; solve all equations with symbols and substitute numbers as the last step. You will need to use the quadratic equation.

B- Calculate the trout's final velocity, that is the VECTOR velocity (magnitude and direction) when the trout splashes.

(It needs to be in meters/second so all numbers need to be converted.)

There is no diagram, I have to draw one. I have no idea how to post it. I don't even know where to start with this one. I just need to be pointed in the right direction PLEASE!

 

[rsk]

Is there are part missing from this question? I assume at this point the trout is dropped and freefalls ? If not, there's no quadratic that I can see.

At the instant the trout is dropped its velocity is ientical to that of the osprey.

OK, first work out the vertical component of this velocity (and horizontal, you'll need it later).

Calcuate how long it will take to travel 12.76 feet at theis vertical velocity.

Use this time to calculate how far she can travel horizontally before hitting the water.

Now, for the final velocity, you need to first calculate the vertical velocity at the time of the splash. The horizontal velocity hasn't changed (if we ignore air resistance). Then combine your two velocity vectors (vector diagram) and find magnitude and angle.

 

[michelleri]

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 give me more info?

 

[rsk]

Ok, first let's breakdown the velocity into horizontal and vertical.

You know the overall velocity is 4.21 mph at 15.32 below the horizontal. Did they give you any conversion factors for mph to m/s? Or do we have to do it from scratch?

 

[michelleri]

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

 

[rsk]

OK, bear with me while I draw it.

 

[michelleri]

No problem! THANKS

 

[rsk]

Ok, here's my basic picture

http://img198.imageshack.us/img198/9063/troutqd8.png

We can breakdown the 1.88 m/s into horizontal and vertical components, using basic trigonometry ie sines and cosines. Have you been shown how to do that?

 

[michelleri]

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?

 

[rsk]

no, I got .497 and 1.813 - it can't possible be bigger than the overall speed (1.88)

I did Uv = 1.88 sin 15.32 and Uh = 1.88 cos 15.32

 

[rsk]

Next step is to calculate the time it takes for the trout to reach the water. For that we need the equations of motion and that's where we're going to need the quadratic.

What equations do you have?

 

[michelleri]

hold on one sec...

 

[rsk]

OK, it's 1am here, I´m going to have to go to bed soon. If no one else can talk you through this I'll try to help you tomorrow.

If you prefer you can PM me your Yahoo messenger ID (if you have one) - it's easier than repeatedly checking for new posts.

 

[rsk]

ok, I'm holding

 

[michelleri]

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.

 

[michelleri]

sent you a pm as well

 

[rsk]

OK, that 2nd one is the one you want to start with. Yf - Yo is the height it's fallen ie 3.89m.

Where they say voy we have said Uv

So what we have is

3.89 = Uvt + 0.5 9.81 t^2

Put in the value we worked out for Uv and we have a quadratic equation to solve for t.

That can be messy - are you allowed fancy calculators or to you have to do it the old fashioned way?

 

[michelleri]

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?

 

[rsk]

Because the horizontal velocity doesn't change at all. SO it's just distance = speed x time, which is really your equation 3

Do a quick websearch for quadratic equation solvers - you just input a, b and c and get the answer. if you're allowed the calculators, why not the web?

 

[michelleri]

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?

 

[rsk]

Yes, assume xo to be zero

 

[michelleri]

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

 

[rsk]

hang on - I don't agree with your value of t

 

[michelleri]

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?

 

[michelleri]

I am hanging on...

 

[rsk]

3.89 = Uvt + 0.5 9.81 t^2

SO
0.5 9.81 t^2 + Uvt - 3.89 = 0 gives a quadratic with a 0 3.89, b = Uov = 0.497 and c = -3.89

 

[rsk]

yes, 1.52 I got as well, good stuff. Just sent you a PM with further steps

 

[michelleri]

AWESOME! Thanks again. Good night.

 

[rsk]

No, in part b you're going to have to calculate a new vertical velocity. It's been accelerating at 9.81 all the way down. Check the PM - it's in there - once you have the vertical velocity you combine it with the horizontal.

Once again, can't guarantee my numbers are exact at this time of night but the method is OK and given that we both get the same thing for the horizontal distance we're likely right than wrong.

Good luck!

 

[michelleri]

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