Physics lab

[Aganack1]

So i did a physics lab this week on Projectile motion. We had a gun (for a lack of a better word) that would shoot a projectile. and we would measure how far it landed from the gun and how far off the ground the gun was to find the time it took to calculate the initial velocity. well the final part of the lab was that the ta gave us a distance and we were to find the angle we needed to shoot at. and he gave us an equation but didn't tell us how he derived it... so now i'm looking to everyone hear to help me understand it. The equation is

Tan(theta) = (Vo)^2/(gR) +/- {[(Vo)^2/(gR)]^2-1+((2(Vo)^2h)/g(R)^2)}^(.5)

Vo is intial Velocity
g is gravity
R is Distance from the gun to the landing spot
h is the heightfrom the floor to the gun
Theta is the angle at witch the gun should be fired


So any help would be perfect

 

[FrustratedUser]

HS physics 101 should give you the answer.

 

[JohnCU]

stairjt with x = poo32sition no
i mean
sttrt witha=celaceration
intrgeate
get v
intrgeate
get x

no god anmit tajhts odesnt work no angles

 

[Heisenberg]

Originally posted by: JohnCU
stairjt with x = poo32sition no
i mean
sttrt witha=celaceration
intrgeate
get v
intrgeate
get x

no god anmit tajhts odesnt work no angles

The only thing worse than being drunk and posting, is being drunk and trying to post the answer to a physics problem.

 

[Gibsons]

Originally posted by: Heisenberg

Originally posted by: JohnCU
stairjt with x = poo32sition no
i mean
sttrt witha=celaceration
intrgeate
get v
intrgeate
get x

no god anmit tajhts odesnt work no angles

The only thing worse than being drunk and posting, is being drunk and trying to post the answer to a physics problem.

great post, though! :beer:

 

[FrustratedUser]

Originally posted by: Heisenberg

Originally posted by: JohnCU
stairjt with x = poo32sition no
i mean
sttrt witha=celaceration
intrgeate
get v
intrgeate
get x

no god anmit tajhts odesnt work no angles

The only thing worse than being drunk and posting, is being drunk and trying to post the answer to a physics problem.

I'd say it's worse to be a fake-drunk trying to immitate a drunk trying to solve a physics problem.

 

[bleeb]

The angle of incidence is equal to the angle of reflection beeyatch!! PWNED!

 

[Aganack1]

can anyone help me?

 

[sonz70]

Do you know how to solve eq's? Do you have any of those values?

 

[franksta]

Originally posted by: bleeb
The angle of incidence is equal to the angle of reflection beeyatch!! PWNED!

Until you exceed the critical angle.

 

[sao123]

Physics = algebra + trig + calc.
ur skrood.

 

[dullard]

Problems like this are extremely simple but have many steps. First thing is a must (despite this week's other physics thread): draw a damn picture.

Start by thinking of the projectile.
[*]It starts with a velocity of Vo at an angle Theta.
[*]What are it's vertical and horizontal velocities?
[*]Vertical: Vy = Vo * sin (Theta)
[*]Horizontal: Vx = Vo * cos (Theta)
That was pretty simple so far.

What happens in the horizontal direction?
[*]I like doing horizontal first since it is easier. But you could look at the vertical direction first.
[*]It will travel for a yet unknown time T.
[*]Ignoring wind resistance, the horizontal velocity, Vx, is constant. Nothing will slow it down in the horizontal direction.
[*]Thus after traveling for a time T, it moves a distance R = Vx*T.
[*]Since you know R, you can solve for T: T = R/Vx.

What happens in the vertical direction?
[*]It too will travel vertically for a time T.
[*]Vertical velocity is not constant this time, due to gravity.
[*]Vertical velocity as a function of time is: Vy(t) = Vy - g*t.
[*]Vertical position as a function of time is: H(t) = h + Vy*t - g*t^2.
[*]At time T, the vertical position is zero (it hits the ground).
[*]Thus: 0 = h + Vy*T - g*T^2.
[*]To keep things simple, solve for T now (quadratic equation):
[*]T = Vy/2g +- [(Vy/2g)^2 +h/g]^0.5

Combine
[*]We know two equations:
[*]T = R/Vx
[*]T = Vy/2g +- [(Vy/2g)^2 +h/g]^0.5
[*]Thus: Vy/2g +- [(Vy/2g)^2 +h/g]^0.5 = R/Vx
[*]Plug in the equations for Vx and Vy.
[*]Solve for Theta and do the rest of your own homework.

 

[Aganack1]

this isn't home work... i was just wondering... I turned my lab report in last night and after i was wondering how he got it and i had tried a few things... to get it by using a few trig identities, but i couldn't get the same answer. i got a quadratic but it wasn't the same... I dont think my ta derived it he just copied it out of his lab manual....

 

[kingers]

What a waste of this poor guy's time. FrustratedUser you have over 18000 posts are all of them as useless as your replies to this post? I understand people have online friends and I'm not here to say that living on a message board is unhealthy or lame, I just don't understand why you must reply to a post when you have nothing useful to say to the poster. If you want to be a wise ass with your "friends" start a new thread. But I suppose I'm preaching to the choir, after 18000+ post you probably tend to forget how to interact on a level higher than that of a high schooler. Perhaps you never knew how and thats what drove you to the message boards. I suppose my message is no more appropriate than that of yours and we all know two wrongs don't make a right, but I feel sorry for Aganack1. Everytime he checks for replies he just gets sandbagged by arrogant "know it alls" which have yet to prove they actually know how to do it. Aganack1 I tried to solve it but I got stuck, I don't think its as "extremely simple" as dullard claims. It looks like you'll have to use some trig identities to solve it, sorry I couldn't help more.