Help me study for my physics test tomorrow!
 

College. Intro to Physics. I'm supposed to be studying right now, but since I have absolutely no background in trig, some steps are a bit "wut". I know how to identify the sides of a right triangle, so that's a check.
I know "soh cah toa" now, so that's also a check.

Unfortunately, I've forgotten (or never got to the point where I understood) what exactly I'm using sine, cosine, tangent, and the inverses for.

I understand that sin = o/h, and cos = a/h, and tan = o/a. Check. But if I have a right triangle and wish to find a side that isn't theta, am I supposed to be solving for one of the variables in o/h, a/h, or o/a? I assume I am, seeing as how if you already have 2 sides, you get the third with P. Theorem.

Now help me put this in perspective. Let's say I have a right triangle whose angle is 20 degrees, and whose opposite side is 10 units. To solve for the adjacent side, I assume I would use the tangent function. So that would be 10/A (where A is the variable of the adjacent side), times the tangent of 20 degrees. Right?

So where do I go from here? I can't rightly multiply a fraction with a variable as the denominator, so do I multiply both sides by the denominator to end up with just 10, and multiply 10 times the tangent of 20 degrees? I would end up with roughly 3.63 as the value of the adjacent side.
Is this correct?

Let me know.


MORE TO COME. for real.

Yes, that is correct. I'm only really good at the trig, so someone else is going to have to tell you about the physics application.

It's been a while, so take my response with a grain of salt. I think that:

1. Setting up the equation

tan(20) = 10 / A

2. Multiply both sides by denominator:

A*tan(20) = 10

3. Divide both sides by tan(20):

A = 10 / tan(20)

4. Simplified:

A = 4.46995109


Eh? We got different answers? Well ignore mine, I haven't taken calculus in forever.

I should've shown my work on paper, Tuna's is right.

I don't get the "divide both sides by the tangent".

You're solving for "A" so you want to isolate the unknown ("A" in this case) on one side so that you can simply punch the other side into the calculator.

Ah, ok. Yea I remember that from Algebra now. :)

Ok. Let me do a problem to make sure I got it, but with the variable in the numerator.

Theta = 45 degrees
Hypoteneus = 12 units.
Find opposite.

Sin(45) = O/12
.70 = O/12
(12).70 = O
O = 8.4

Is that right?

Correct.
EDIT: If you understand the principle of sohcahtoa, you'll be able to apply that correctly, it's just the Algebra where most people screw up the solution.

Word. Ok. So I have the basics. Now, when doing 2 dimensional kinematics, that trig comes into play, but I also have to remember some formulas. I can put those on my phone, so I'm not worried too much about it, since my phone is also my calculator, which is allowed.

So, how would you guys go about solving something like this (taken from a homework problem):

A projectile is launched from and returns to ground level. Air resistance is absent. The horizontal range of the projectile is R = 175 m, and the horizontal component of the launch velocity is Vox = 25 m/s. Find the vertical component Voy of the launch velocity.

Ok a little heads up I noticed with the dumb google calculator

tan(20 degrees) is actually equal to 0.36, not sure why google calculator was giving me that other number. But the steps are the same. So the first answer is actually something like 27.

And yes you're answer looks right

Oh, one more thing about trig real quick. If I want to find the angle of a right triangle, I know I need to use inverse sin, cos, or tan.

Walk me through the steps of finding the angle of a right triangle whose hypoteneus is 5 units, and whose opposite side is 7 units.

opposite side can't be larger than the hypotenuse :)
. . .but if the numbers are reversed, you have

sin(x) = 5/7 = .7

x = arcsin (.7)

I think it's as simple as arcsin(7/5). Just put that into your calculator.

It's basically just the reverse, there's no steps involved, unless it was values used on 45-45-90 triangles or 30-60-90

Plus what skaltezon said LOL

K. Disregarding the fact that my triangle is breaking all the rules, I would simply divide the opposite side by the hypoteneus and multiply that number by the inverse of sin? That works for me.

Hypotenuse is always smaller than the opposite. If I recall correctly, you can plug in arcsin(7/5) in a calculator to get your answer.

edit: damn, I was too slow :(

I haven't taken college physics but isn't the equation just displacement = velocity * time ?

Yes. That will give you the angle.

Well displacement is actually velocity over time (d = v/t) if I remember correctly, but that's one dimensional. The problem has both the x and y dimension, and is asking for the components of the vector, which the vector would be a diagonal line. I would need to figure out the x component and the y component that made the resultant vector.

No. Find the value of 5/7 and look in a table to see what angle has that sine value. Or use your scientific calculator, if that's permitted.

Nevermind, that's acceleration.

I think Tuna's right with d = v * t .

Yea, I was thinking of Acceleration.

Well, the time from firing to impact is 175/25 = 7 secs.

Reaches apogee midway at 3.5 secs., where vertical velocity = 0.

What velocity will be exactly cancelled by gravity after 3.5 secs.?

I'm guessing (10meters/sec/sec)(3.5 sec.) = 35 meters per second.

Now you can calculate muzzle velocity. :)

Alright man, I looked up 2 dimensional kinematics and I'm still not sure how to solve this problem. My best guess:

Using the equation: y=v*t+.5*a*t^2

where:
y = vertical displacement
v= velocity
t= time
a = acceleration

and figuring out that t=7

you have:

0 = v*7+.5*(-9.8)*7^2

which got me

v= 34.3 m/s

have no idea if that's actually right though

Ah cool just noticed you got the same answer, guess we're on the same page.

Ah yes, that makes sense.
Seeing the equation written that way (instead of the way my professor wrote it) makes things a lot easier. It's just a matter of plug and play then.

Yeah which equation you use is really all dependent on what you're given in the first place.

Here's the list I got it from in case you don't have them all:

Horizontally Launched Projectile Problems

There are 4 of them that he said we'd need. Those I have typed into my phone. I don't have any with like Viy though... does that mean the initial velocity of Y, or is that the initial velocity multiplied by a number represented by Y?

Just initial velocity of Y, basically the 34.3 that you solved for in the other problem.

K. Cool.
I think that will be it for now. It's our first test, and most of it will be basic, like converting units and stuff like that. I just wanted to be sure I had a grasp on the kinematics and the trig stuff. I have the formulas now, so I should (I hope, if I don't choke) do OK.

Thanks guys.

I have a test in Biochemistry tomorrow...so we'll both be having fun.

Good luck to you, sir.

Don't forget the critical formula that tells you which formula to use. http://i485.photobucket.com/albums/r...de_999/doh.gif

Yeah I got a history test Thursday. Have fun studying guyz!

Calculus test here. Damn is tomorrow national test day?