1. Velocity of the first particle at maximum height

A projectile is launched with speed $u$ at angle $\theta=\pi/3$.

Its initial components are: $u_x=u\cos\frac{\pi}{3}=\frac{u}{2}, \qquad u_y=u\sin\frac{\pi}{3}=\frac{\sqrt{3}u}{2}$

At maximum height, vertical velocity becomes zero, so the velocity of the first particle is: $\vec v_1=\frac{u}{2}\,\hat i$

The second particle has velocity: $\vec v_2=u\hat i$

2. Velocity just after completely inelastic collision

Both particles have equal mass $m$, and they stick together.

Using conservation of linear momentum in horizontal direction: $m\left(\frac{u}{2}\right)+m(u)=(2m)V$

So, $V=\frac{\frac{u}{2}+u}{2}=\frac{3u}{4}$

Thus the combined mass moves horizontally with speed: $V=\frac{3u}{4}$

There is no vertical component immediately after collision, since both vertical components are zero.

3. Height at which collision occurs

The collision occurs at the maximum height of the projectile.

Maximum height is: $H=\frac{u^2\sin^2\theta}{2g}$

With $\theta=\pi/3$, $\sin^2\frac{\pi}{3}=\frac{3}{4}$

Hence, $H=\frac{u^2\cdot \frac{3}{4}}{2g}=\frac{3u^2}{8g}$

4. Time taken by combined mass to reach ground

After collision, the combined mass starts from height $H$ with zero vertical velocity.

So its fall time is: $H=\frac{1}{2}gt^2$

Therefore, $t=\sqrt{\frac{2H}{g}}=\sqrt{\frac{2}{g}\cdot \frac{3u^2}{8g}}$ $t=\sqrt{\frac{3u^2}{4g^2}}=\frac{\sqrt{3}u}{2g}$

5. Horizontal distance covered after collision

Horizontal distance covered by the combined mass before hitting ground: $x=Vt$

Substitute $V=\frac{3u}{4}$ and $t=\frac{\sqrt{3}u}{2g}$: $x=\frac{3u}{4}\cdot \frac{\sqrt{3}u}{2g}$ $x=\frac{3\sqrt{3}}{8}\frac{u^2}{g}$

6. Compare with options

This matches: $\boxed{\frac{3\sqrt{3}}{8}\frac{u^2}{g}}$

So the correct option is B.

7. Verification with stored answer

Stored correct answer: B

Our derived answer: B

They agree.