Physics
posted by Bob on .
The problem contains graphics that I don't know how to show other than in an image format:
h t t p : / / b i t . l y / c D H P V M
h t t p : / / b i t . l y / 9 m b y C R

And I still haven't gotten an answer for my previous question here:
h t t p : / / b i t . l y / d g 6 o O F 
I can't make the "previous" graphic open.
On this problem, label the horizontal distance H (dcosPHI) and the final height yf(dsinPHI)
horizontal:
VcosTheta*time=x=dcosPHI
vertical
yf=VsinTheta*time1/2 g t^2
from the first equation
time=dcosPHI/VcosTheta
and putting this in the second equation
dsinPHI=VsinTheta*dcosPHI/VcosTheta1/2 g (dcosPHI/VcosTheta)^2
multiplying both sides by (VcosTheta)^2
dsinPHI(VcosTheta)^2=V^2 sinThetacostheta d cosPHI=1/2 g dcosPHI
so take the derivative of d with respect to theta, set to zero, solve for theta in terms of V and PHI.
A bit of algebra is required, so get a large pad of paper. 
What does d= ?
(Not sure what to take the derivative of) 
d is the height up the hill, as drawn on your figure. As I recall, you are trying to maximize d, so d d/dTheta=0=..