A

Step 1: The particle's velocity at any instant is given as a vector with magnitude $u$ making an angle $\alpha$ with the horizontal. Therefore, the horizontal and vertical components of the velocity can be expressed as:

• $u_x = u \cos(\alpha)$ (horizontal component)
• $u_y = u \sin(\alpha)$ (vertical component)

Step 2: To find the time when the particle will be moving at a right angle to its initial direction, we need the resultant velocity vector to have a horizontal direction (i.e., only the horizontal component should remain). Therefore, we need to consider the effect of gravity on the vertical motion. The vertical velocity will change under uniform acceleration due to gravity ($g$) as:
$$ v_y = u_y - g t = u \sin(\alpha) - g t $$
For the particle to move horizontally ($v_y = 0$), we set:
$$ u \sin(\alpha) - g t = 0 $$
Step 3: Solving for $t$ gives us:
$$ g t = u \sin(\alpha) $$
$$ t = \frac{u \sin(\alpha)}{g} $$
Conclusion: Therefore, the time after which the particle will be moving at a right angle to its initial direction is $t = \frac{u \sin(\alpha)}{g}$. Hence, the correct answer is option A.