A particle is projected with velocity $V $,at an angle of $75^0$ to the horizon,from the foot of a plane whose inclination is $30^0.$I have to find where it will strike the plane.
In the given problem we have to find the Range $R$ corresponding to the inclined plane with angle of inclination $\theta$.
For inclined plane we have ,Range $R=\frac{2u^2\sin(\alpha)\cos(\alpha+\theta)}{g\cos^2(\theta)}$ where $u:=$Velocity of the projectile;$\alpha:=$Angle of projectile corresponding to the inclined plane;$\theta:=$Angle of inclination of plane with horizon.
According to the problem,we have $\theta+\alpha=75^0;\theta=30^0$
So,$R=\frac{2u^2\sin(45^0)\cos(75^0)}{g\cos^2(30)}=\frac{2V^2\times \frac{1}{\sqrt2}\times\frac{\sqrt3-1}{2\sqrt2}}{9.8\times (\frac{3}{4})}=\frac{V^2(\sqrt3-1)}{14.7}$
But,the correct answer is $R=\frac{V^2(\sqrt3-1)}{48}$. I think I have made a mistake while assigning the value to the variables...please give it a glance over the approach in particular to the value of acceleration due to gravity