Posted by **Bob** on Tuesday, September 7, 2010 at 9:48pm.

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

- Physics -
**URGENT**, Tuesday, September 7, 2010 at 9:57pm
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

- Physics -
**bobpursley**, Tuesday, September 7, 2010 at 10:45pm
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*time-1/2 g t^2

from the first equation

time=dcosPHI/VcosTheta

and putting this in the second equation

dsinPHI=VsinTheta*dcosPHI/VcosTheta-1/2 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.

- Physics -
**URGENT**, Tuesday, September 7, 2010 at 11:02pm
What does d= ?

(Not sure what to take the derivative of)

- Physics -
**bobpursley**, Wednesday, September 8, 2010 at 12:49pm
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=..

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