1. Velocity of the first particle at the highest point

A particle is projected with speed $u_0$ at angle $\alpha$.

Its horizontal and vertical components initially are: $u_x=u_0\cos\alpha, \qquad u_y=u_0\sin\alpha$

At the highest point of projectile motion, the vertical velocity becomes zero, so the velocity of the first particle is purely horizontal: $\vec v_1=(u_0\cos\alpha)\,\hat i$

2. Velocity of the second particle when it collides

The second identical particle is thrown vertically upward from the ground with speed $u_0$.

We need its velocity at the instant of collision.

The collision occurs at the highest point of the first particle.

Time taken by the first particle to reach highest point: $t=\frac{u_0\sin\alpha}{g}$

At this same time, the vertical velocity of the second particle is: $v_2=u_0-gt$

Substitute $t$: $v_2=u_0-g\left(\frac{u_0\sin\alpha}{g}\right)=u_0(1-\sin\alpha)$

So its velocity is: $\vec v_2=u_0(1-\sin\alpha)\,\hat j$

3. Check that collision is possible

Height of highest point of first particle: $H=\frac{u_0^2\sin^2\alpha}{2g}$

Height of second particle at time $t=\frac{u_0\sin\alpha}{g}$: $y=u_0 t-\frac12 gt^2$ $y=u_0\left(\frac{u_0\sin\alpha}{g}\right)-\frac12 g\left(\frac{u_0^2\sin^2\alpha}{g^2}\right)$ $y=\frac{u_0^2}{g}\sin\alpha-\frac{u_0^2}{2g}\sin^2\alpha$

For collision at the highest point, this must equal $H$: $\frac{u_0^2}{g}\sin\alpha-\frac{u_0^2}{2g}\sin^2\alpha=\frac{u_0^2\sin^2\alpha}{2g}$

This gives: $\sin\alpha=\sin^2\alpha$ $\sin\alpha(1-\sin\alpha)=0$

Thus, for a nontrivial collision, $\sin\alpha=1 \Rightarrow \alpha=\frac\pi2$

But in standard momentum-based JEE treatment, the intended interpretation is that at collision instant the second particle is moving vertically upward and the first is moving horizontally, and we use momentum conservation directly. Let us proceed with that intended setup.

4. Apply conservation of momentum in completely inelastic collision

Masses are identical, each of mass $m$.

Before collision, total momentum is: $\vec p=m\vec v_1+m\vec v_2$

So, $p_x=m u_0\cos\alpha$ $p_y=m u_0(1-\sin\alpha)$

After collision, the particles stick together, so total mass becomes $2m$. Let the composite move at angle $\theta$ with the horizontal.

Then, $\tan\theta=\frac{p_y}{p_x}=\frac{u_0(1-\sin\alpha)}{u_0\cos\alpha}$ $\tan\theta=\frac{1-\sin\alpha}{\cos\alpha}$

Now simplify: $\frac{1-\sin\alpha}{\cos\alpha}=\frac{(1-\sin\alpha)(1+\sin\alpha)}{\cos\alpha(1+\sin\alpha)}$ $=\frac{1-\sin^2\alpha}{\cos\alpha(1+\sin\alpha)}$ $=\frac{\cos^2\alpha}{\cos\alpha(1+\sin\alpha)}$ $=\frac{\cos\alpha}{1+\sin\alpha}$

Using the identity: $\tan\left(\frac\pi4-\frac\alpha2\right)=\frac{1-\sin\alpha}{\cos\alpha}$

Therefore, $\theta=\frac\pi4-\frac\alpha2$

This result does not match any option.

5. Compare with given options and stored answer

From correct momentum analysis, the angle should be: $\boxed{\theta=\frac\pi4-\frac\alpha2}$

This is not among options A, B, C, D.

Hence the stored correct answer $A$ is not consistent with the physics of the problem as stated.