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)
yf=VsinTheta*time-1/2 g t^2
from the first equation
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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