1my bad i did a silly mistake
it is
1+\tan(2(\theta + \phi))\tan\phi=0
shubomoys is wrong
dont know where i am making the mistake
but i am getting something like this
1+\tan(\theta +\phi)\tan(\phi)=0
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8 Answers
49instead of what u got i think u should get,
1-tan\left(\theta +\phi \right)tan\left( \phi\right) =0
dividing both sides by tan\left(\theta +\phi \right)+tan\left( \phi\right)
giving tan\left(\theta +2\phi \right)=tan(\pi /2)
thus the result
1
23let
\hat{x},\hat{y}=unit vectors along the incline and perpendicular to the incline
\vec{g}=gsin\phi (x)-gsin\phi (y)
v_{x}=vcos(\theta +\phi )
v_{y}=vsin(\theta +\phi )
time of flight = t
t=2v_{y}/gcos(\phi ) =\frac{vsin(\theta +\phi )}{gcos(\phi )}
Range=R
R=v_{x}t+\frac{gsin(\phi) t^{2}}{2}
on rearranging R=\frac{2v^{2}sin(\theta +\phi)cos\theta }{gcos^{2}(\phi )}
\frac{dR}{d\theta }=0 ,\Rightarrow cos(2\theta +\phi )=0