Projectile Motion problem halp

[Heavencloud]

So for Physics lab we had this extra credit part of the lab where they wanted us to convert

X=Vcos(θ)[((-Vsinθ)±√(Vsinθ^2-(2gh))/-g))]

Where V = initial velocity
X = Range
h = height launched
g = gravity constant (9.8m/s)

Anyway, none of that is really important except for the equation that it goes into, which is

tan θ = V^2/gX ± [((V^2)/gX)^2 - 1 + ((2V^2)*h)/gX^2]^1/2

Maybe I'm just too rusty of my algebra and trig but I can't seem to figure out how they got from the first equation to the second. The reason that we had to put it into the form of the first equation was so that we could use the quadratic formula to solve for the distance that we launched our projectile. I know it's kinda hard to read those equations, wish I had maple but this is even how they gave em to us in our lab... FUUU.

Anyway, if anyone has any ideas or could get me started I'd greatly appreciate it.

[Eliseos]

You can simplify sin^2 to 1/2 (1-cos(2 x)). Then to get tan you'll have to work the equation to get sin/cos on one side. That's how I would start it anyway, if I get some time I'll try to sit down and work through it. You might also need to square both sides to get rid of the root, which would explain how they get to V^2.

EDIT: Also, you have an unbalanced parenthesis. There is an extra ) at the end, which seems to be paired with the []'s?

[Woozie]

Is theta the angle at which it was launched or is it the polar angle of the position vector at any given time?

[Woozie]

Nevermind, it's definitely the initial angle. Now, under the square root you have "-2gh". Are you sure that isn't supposed to be +2gh?

[Kaylia]

Nevermind, it's definitely the initial angle. Now, under the square root you have "-2gh". Are you sure that isn't supposed to be +2gh?

Same thing. It change with your axis.

[Heavencloud]

Is theta the angle at which it was launched or is it the polar angle of the position vector at any given time?

Angle at which it was launched, sorry I forgot to add that in.

Basically it looked like this:

http://www.ux1.eiu.edu/~cfadd/1150/0...ages/proj1.gif

And I just double checked and it was +2gh, misread it >.>

[Heavencloud]

You can simplify sin^2 to 1/2 (1-cos(2 x)). Then to get tan you'll have to work the equation to get sin/cos on one side. That's how I would start it anyway, if I get some time I'll try to sit down and work through it. You might also need to square both sides to get rid of the root, which would explain how they get to V^2.

EDIT: Also, you have an unbalanced parenthesis. There is an extra ) at the end, which seems to be paired with the []'s?

My bad, ugh haha this is why I love using Maple, I should just DL it so I can write out equations like this with a better view.
It should've been, although it's practically the same. fixed the +2gh aswell.

X=Vcos(θ)[(-Vsinθ)±√(Vsinθ^2+(2gh))/-g)]

[Eliseos]

Can try this, it looks kind of bad with the black forum skin but it's better than nothing http://www.codecogs.com/latex/install.php

[Woozie]

I'm not getting the right answer and I don't really have time to work on the problem. I'm getting tan in terms of the same variables that you have them in terms of, so I'm guessing it's going to take a bunch of algebra to get my answer to look like yours. My answer is much bigger and much more complicated.

What I did was to write x(t) = v*cosθ*t, y(t) = h + v*sinθ*t-1/2*g*t^2. In the second equation, I solve for v*sinθ*t, and divide this answer by v*cosθ*t. The v and the t cancel and I'm left with sinθ/cosθ, which is just tanθ. So this is equal to y(t)-h+1/2*g*t^2/x(t)

Now we plug in t = <the thing you had for x> divided by v*cosθ. So, just the part of x that doesn't have the v*cosθ. This gives you y(t)=0 automatically, and x(t) = X. After making these substitutions, you end up with an answer that's similar to yours and dimensionally consistent, but still depends on sinθ. I use the equation you had (x=blah blah blah), and rewrite the cos in terms of sinθ and solve for it, and substitute that for every sinθ in my equation. This gives a really long expression, but it's tanθ expressed in terms of the same variables as your answer. I don't know how to get from what I have to what you have. If I have more time I'll work on it tomorrow.

[Heavencloud]

I'm not getting the right answer and I don't really have time to work on the problem. I'm getting tan in terms of the same variables that you have them in terms of, so I'm guessing it's going to take a bunch of algebra to get my answer to look like yours. My answer is much bigger and much more complicated.

What I did was to write x(t) = v*cosθ*t, y(t) = h + v*sinθ*t-1/2*g*t^2. In the second equation, I solve for v*sinθ*t, and divide this answer by v*cosθ*t. The v and the t cancel and I'm left with sinθ/cosθ, which is just tanθ. So this is equal to y(t)-h+1/2*g*t^2/x(t)

Now we plug in t = <the thing you had for x> divided by v*cosθ. So, just the part of x that doesn't have the v*cosθ. This gives you y(t)=0 automatically, and x(t) = X. After making these substitutions, you end up with an answer that's similar to yours and dimensionally consistent, but still depends on sinθ. I use the equation you had (x=blah blah blah), and rewrite the cos in terms of sinθ and solve for it, and substitute that for every sinθ in my equation. This gives a really long expression, but it's tanθ expressed in terms of the same variables as your answer. I don't know how to get from what I have to what you have. If I have more time I'll work on it tomorrow.

Wow that's crazy, I can't believe they're putting something like this for my third lab in into physics lol. If it helps the equations that we started with were:

h=vsinθ-1/2gt^2 and X=v*cosθt which is what we derived the X=Vcos(θ)[(-Vsinθ)±√(Vsinθ^2+(2gh))/-g)] from. Maybe there is a better way to get to it just using these since they're so much simpler?

[Melchiah]

I don't know that I fully understand what you're trying to communicate. Are both equations given to you, or do you have to derive the second from the first or what? The first equation you listed appears to be a position function. The first derivative of a position function yields the velocity function, and the second derivative yields the acceleration function. Right now, you have two equations in the OP, meaning you can solve for a maximum of two variables. I haven't tried substituting the first equation into all the X-values in the second yet, but that's all I see that can be done with what is provided.

[Heavencloud]

I don't know that I fully understand what you're trying to communicate. Are both equations given to you, or do you have to derive the second from the first or what? The first equation you listed appears to be a position function. The first derivative of a position function yields the velocity function, and the second derivative yields the acceleration function. Right now, you have two equations in the OP, meaning you can solve for a maximum of two variables. I haven't tried substituting the first equation into all the X-values in the second yet, but that's all I see that can be done with what is provided.

We were given h=vsinθ-1/2gt^2 and X=v*cosθt in the beginning which we used to find the X=Vcos(θ)[(-Vsinθ)±√(Vsinθ^2+(2gh))/-g)]. The tan θ = V^2/gX ± [((V^2)/gX)^2 - 1 + ((2V^2)*h)/gX^2]^1/2 was given but I think the way to get to the tan theta equation is through the initial two equations, not the one that we found with those two equations.