# Angle of projection above an incline?

by Buck268
Tags: angle, incline, projection
 P: 333 Just apply $$s = ut + \frac{1}{2}at^2$$ 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.
 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... $$s = VoCos\gamma [\frac{2VoSin\gamma}{g} * Sec \theta] - \frac{gSin \theta}{2} * [\frac{2VoSin\gamma}{g} * Sec \theta]^2$$ Which is in the form $$s = ut + \frac{1}{2}at^2$$ 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 $$t = \frac{2Voy}{g}*Sec\theta$$ where naturally $$Voy = VoSin\gamma$$...
 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: $$r(t) = \frac{2Vo^2 * Sin\gamma * Cos\gamma}{g*Cos\theta} - \frac{2Vo^2 * Sin^2 \gamma}{gCos\theta}*Tan\theta$$ hmmmm...
 P: 9 Ahhh well I've came up with $$\gamma = \frac{1}{2}ArcCot(Tan \theta)$$