Question

for a projectile thrown up an incline plane (making an angle $\theta$ with
horiontal), maximum range on the incline
can be achieved by projecting at an angle .. from the horizontal ( $\theta$ is acute).

Answer 1

In $y-direction,$

let $t$ be the times of height.

Then,

$0=\left(v\sin \alpha \right)t-\frac{gcos\theta t^2}{2}$

$t=\frac{2v\sin \alpha }{g\cos \theta }$

So, in $x-direction$

$R=v\cos \alpha \left[2\left(\frac{v\sin \alpha }{g\cos \theta }\right)\right]-\frac{g\sin \theta }{2}\left[\frac{2v\sin \alpha }{g\cos \theta }\right]$

$R=v\left[\frac{2v\sin \alpha -\cos \theta }{g\cos \theta }\right]-\frac{2v^2\sin \theta -{\sin }^2\alpha }{g{\cos }^2\theta }$

$R=\frac{2v^2\sin \alpha }{g{\cos }^2\theta }(\cos \alpha .\cos \theta -\sin \theta .\sin \alpha )$

$R=\frac{2v^2\sin \alpha }{g\cos \theta }.\left(\cos \right(\alpha +\theta \left)\right)$

For $R$ to be max,

Applying $AM-GM$ in equality,

So, $\cos (\alpha +\theta )=\sin \alpha$

$\Rightarrow \cos \alpha .\cos \theta -\sin \alpha -\cos \theta =\sin \alpha$

$\Rightarrow \cos \alpha \cos \theta =\sin \alpha (1+\cos \theta )$

$\Rightarrow \tan \alpha =\frac{\cos \theta }{1+\cos \theta }$

$\alpha ={\tan }^{-1}\left[\frac{\cos \theta }{1+\cos \theta }\right]$

So, $R=2v^2\sin \alpha g{\cos }^2\theta .\left(\cos \right(\alpha +\theta \left)\right)=\frac{2v^2{\sin }^2\alpha }{g{\cos }^2\theta }$

$R=\frac{2v^2}{g{\cos }^2\theta }\left[{\sin }^2\left(ta{n}^{-1}\left[\frac{\cos \theta }{1+\cos \theta }\right]\right)\right]$