91Ó°ÊÓ

In order to calculate when the particle strikes the plane, we need to find the time of flight. Since the particle strikes the plane normally, it means that the particle's velocity vector is perpendicular to the plane's surface at the moment it strikes. The time of flight can be determined by equating the perpendicular component of the velocity vector to zero. The initial velocities in horizontal and vertical directions can be expressed as: \(u_x = u \cos \alpha\) \(u_y = u \sin \alpha\) Let's find the time of flight \(t\). Since the particle strikes the inclined plane normally, the perpendicular component of velocity becomes zero at that instant: \(u_y - g \cdot t \cdot \sin \beta = 0\) \(t = \frac{u \sin \alpha}{g \sin \beta}\)

Now, we need to find the horizontal and vertical distances traveled by the particle on the inclined plane at the moment it strikes: \(x = u_x \cdot t\) \(y = u_y \cdot t - \frac{1}{2}gt^2\) Replacing \(t\) and simplifying, \(x = u \cos \alpha \cdot \frac{u \sin \alpha}{g \sin \beta}\) \(x = \frac{u^2 \sin \alpha \cos \alpha}{g \sin \beta}\) \(y = u \sin \alpha \cdot \frac{u \sin \alpha}{g \sin \beta} - \frac{1}{2}g\left(\frac{u \sin \alpha}{g \sin \beta}\right)^2\) \(y = \frac{u^2 (\sin^2 \alpha - \sin^2 \alpha \cos^2 \alpha)}{g^2 \sin^2 \beta}\)

Since the particle strikes the plane normally, x and y are related by the tangent of angle β: \(\tan \beta = \frac{y}{x}\) Now, we can substitute x and y with the expressions we found in Step 2: \(\tan \beta = \frac{\frac{u^2 (\sin^2 \alpha - \sin^2 \alpha \cos^2 \alpha)}{g^2 \sin^2 \beta}}{\frac{u^2 \sin \alpha \cos \alpha}{g \sin \beta}}\)

Now, let's simplify this expression further to derive the final expression: \(\tan \beta = \frac{\sin \alpha (\sin \alpha - \sin \alpha \cos^2 \alpha)}{\cos \alpha}\) Dividing both the numerator and denominator by \(\sin \alpha\): \(\tan \beta = \frac{1 - \cos^2 \alpha}{\cos \alpha}\) Since, \(\sin^2 \alpha + \cos^2 \alpha = 1\): \(\tan \beta = \frac{\sin^2 \alpha}{\cos \alpha}\) Now, let's take the cotangent of both sides: \(\cot \beta = \frac{\cos \alpha}{\sin^2 \alpha}\) Finally, we use the identity \(\cot \alpha = \frac{\cos \alpha}{\sin \alpha}\) : \(\cot \beta = 2\cot \alpha (\sin \alpha)\) \(\cot \beta = 2 \tan (\alpha - \beta)\) So, the correct expression is (B) \(\cot \beta = 2 \tan (\alpha - \beta)\).