Optics + Projectile motion + Relative motion

OK, the problem is a bit tricky. But first I will discuss the solution.

The image will appear relative to the ball when the velocity is perpendicular to the mirror. At the beginning, the Ball's velocity makes $60^0$ with the horizontal while the mirror makes $30^0$ . And that's why the ball's image will seem to be moving relative to the ball. But when the velocity makes $30^0$ with the ground, for a moment, only for a moment, their relative velocity will be zero. Because, When you walk parallel to your dressing table mirror, you see your image still, don't you?

Now, the calculations. We have, $v_0=20m/s$ $\theta_b=60^0$ $\theta_m=30^0$ $g=10m/s^2$

When velocity makes $30^0$ , Assume, Vertical component =v

Horizontal component = $v_0 cos\theta_b$ $tan \theta_m=\frac{v}{v_0cos\theta_b}$ $or, v=v_0cos\theta_b tan \theta_m$

For this vertical component again, $v_0sin\theta_b-v=gt$ $or, t=\frac{v_0(sin\theta_b-cos\theta_btan\theta_m)}{g}$ $=\frac{10(\frac{\sqrt{3}}{2}-\frac{1}{2}\frac{1}{\sqrt{3}})}{10}s$ $=\frac{1}{2}(\sqrt{3}-\frac{1}{2\sqrt{3}})s$

Final job, in the $\frac{a}{\sqrt{b}}$ form. $=\frac{1}{2}(\frac{3+1}{\sqrt{3}})$ $=\frac{2}{\sqrt{3}}$

Thus, $a+b=2+3=\boxed{5}$

You got it!