A particle is projected at an angle $\alpha$ with the horizontal from foot of the plane, whose inclination to horizontal is $\beta$. Show that it will strike plane at right angles if $ cot \beta = 2* tan(\alpha - \beta)$
I know that time $T$ and range $R$ for a projectile on an inclined plane is given by $T = \dfrac{2*u*sin(\alpha - \beta)}{g*cos\beta}$ and $R=\dfrac{2*u^2*cos(\alpha)*sin(\alpha - \beta)}{g*cos^2\beta}$
If I use the equation $v = u + a*t$ where $v$ = final velociy, $u$ = initial velocity, $a$ = acclereation and $t$ = time,
then along the incline, we have $v=0$, $u$ = $u*cos(\alpha-\beta)$,$a=-g*sin\beta$ and $t=T$ and we get the desired result.
But if I use the relation $v^2 = u^2 + 2*a*S$ where $v=0$, $u$ = $u*cos(\alpha-\beta)$ ,$a=-g*sin\beta$ and $s=R$ then I get a different result. Why is it so? Is it wrong to assume that equation $v^2 = u^2 + 2*a*S$ is true in this case?
It is not different. You get $$\tag1 \cos^2\beta\cos^2(\alpha-\beta) = 4\sin\beta\sin(\alpha-\beta)\cos\alpha $$ To make it easier to handle, let $\gamma = \alpha - \beta$ and expand $$ \cos\alpha = \cos(\beta + \gamma) = \cos\beta\cos\gamma -\sin\beta\sin\gamma $$ Substitute that into (1) and move everything on the left: $$ (\cos\beta\cos\gamma)^2 - 4\sin\beta\sin\gamma\cos\beta\cos\gamma + 4(\sin\beta\sin\gamma)^2 = 0$$ Does that look familiar? (Think $(A-B)^2$ with suitable $A$ and $B$.)
It is natural that when using $v^2 = u^2 + 2aS$ you should get a more complicated result - one that has been effectively squared, since $v^2 = u^2 + 2aS$ is essentially a squared version of $v = u + at$.
In the future, I would suggest that when writing vector components, you use some subscript to indicate that, such as $u_x$ or $u_\parallel$. Otherwise it gets confusing whether by $u$ you mean the magnitude of the initial velocity, or the initial velocity in a particular direction - you seem to be switching between the two meanings and end up writing absurdities such as $u = u \cos(\alpha -\beta)$.