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Angle of projection above an incline?

by Buck268
Tags: angle, incline, projection
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Buck268
#1
Feb7-05, 08:36 AM
P: 9
Well, right now I'm working on one helluva a problem... Basically, a projectile is given a velocity of V sub "o" (Vo). The launch angle is gamma degrees above an surfaced which is inclined theta degrees above the horizontal. I'm tasked with finding its range along the inclined surface as well as finding the optimal angle gamma to maximize the range.

So far, what I've done is rotate the coordinate system suh that the x-axis is along the inclined surface with the origin of the (x,y)-axis being the intersection of this inclined surface, the ground, and the initial launch point.

This provides for the following components of "G" (which I'm taking to be -9.8m/s^2). Gx = -g*Sin Theta and Gy = -g*Cos Theta. This took a lil goemetry to obtain (had to draw a couple diagrams in order to work it all out).

Then I solved for t = (2Voy/g)Sec theta as well as Vox = VoxCos(gamma + theta). Of coarse, I now see an error, as that Vox would be for (x,y) with respect to the ground, not the incline surface... I'll post where my correction has lead to in a second...
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Gamma
#2
Feb7-05, 09:32 AM
Gamma's Avatar
P: 333
Just apply

[tex] s = ut + \frac{1}{2}at^2 [/tex]

perpendicular to the surface and along the surface. Use the fact that perpendicular to the surface s = 0.

Always a clear diagram with angles and directions is usful to quickly solve this kind of problems.
Buck268
#3
Feb7-05, 09:48 AM
P: 9
OK, well I've never seen that form, but I'm using R(t) = do + Vox*t - .5Axt^2. Looks to me like s = r(t), u = Vox, and of course Do = 0...

Anyways, The problem I'm running into is simplifiing to a deferentiable form, since I'm trying to find the maximum gamma (probably should solve for gamma first, too?). The equation I've worked out seems to be correct, but I have trig functions of both gamma and theta. I'll see if I got AutoCAD laying around, if so I'll make a lil drawing right quick, but lemme show you the equation I have...

[tex]s = VoCos\gamma [\frac{2VoSin\gamma}{g} * Sec \theta] - \frac{gSin \theta}{2} * [\frac{2VoSin\gamma}{g} * Sec \theta]^2[/tex]

Which is in the form [tex] s = ut + \frac{1}{2}at^2 [/tex]

Like I said, I'm not sure how/if this is differentiable or solvable for gamma... I suppose this equation is the equation for the range of the projectile along the inclined surface, but I would still have to zero out the derivitive with respect to gamma in order to find the gamma which provides for max range, correct?

Those "g" vectors (gravity... 9.8m/s^2) came from the formula I used for time, which worked out to be [tex]t = \frac{2Voy}{g}*Sec\theta[/tex] where naturally [tex]Voy = VoSin\gamma[/tex]...

Buck268
#4
Feb7-05, 10:10 AM
P: 9
Angle of projection above an incline?

the equation for the range (r(t) or s if you prefere) seems to work out to:
[tex]r(t) = \frac{2Vo^2 * Sin\gamma * Cos\gamma}{g*Cos\theta} - \frac{2Vo^2 * Sin^2 \gamma}{gCos\theta}*Tan\theta[/tex]

hmmmm...
Buck268
#5
Feb7-05, 10:16 AM
P: 9
Ahhh I'm retared... I think that everything is supposed to be a constant except gamma... Maybe that would make it easier
Buck268
#6
Feb7-05, 10:48 AM
P: 9
Ahhh well I've came up with
[tex]\gamma = \frac{1}{2}ArcCot(Tan \theta)[/tex]
Gamma
#7
Feb7-05, 01:03 PM
Gamma's Avatar
P: 333
That's what I got too.
Buck268
#8
Feb7-05, 01:40 PM
P: 9
Excellent, thank you for the assistance

...now on to providing a general proof that for a given speed there are two angles between 0 and 90 degree which have the same range... Yay...


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