A particle of mass $m$ is projected from the ground with an initial speed $u_0$ at an angle $\alpha$ with the horizontal. At the highest point of its trajectory, it makes a completely inelastic collision with another identical particle, which was thrown vertically upward from the ground with the same initial speed $u_0$. The angle that the composite system makes with the horizontal immediately after the collision is

Step 1: Find the velocity of the first particle at the highest point of its trajectory.

At the highest point, the vertical component of the velocity of the first particle is zero, leaving only the horizontal component.

$$v_{1x} = u_0 \cos\alpha$$ $$v_{1y} = 0$$

Step 2: Find the velocity of the second particle at the collision height.

Using the conservation of mechanical energy, the speed of the second particle at height $H$ is determined by its initial speed $u_0$.

$$\frac{1}{2}m u_0^2 = \frac{1}{2}m v_2^2 + mgH$$

Since the first particle is also at height $H$ at the apex of its motion:

$$\frac{1}{2}m u_0^2 = \frac{1}{2}m (u_0 \cos\alpha)^2 + mgH$$

Equating the two energy expressions yields the vertical velocity of the second particle.

$$v_2 = u_0 \cos\alpha$$ $$v_{2x} = 0$$ $$v_{2y} = u_0 \cos\alpha$$

Step 3: Apply conservation of momentum for the inelastic collision.

The two identical particles of mass $m$ coalesce into a composite system of mass $2m$.

$$m v_{1x} = (2m) v_x$$ $$v_x = \frac{u_0 \cos\alpha}{2}$$ $$m v_{2y} = (2m) v_y$$ $$v_y = \frac{u_0 \cos\alpha}{2}$$

Step 4: Calculate the angle of the composite system with the horizontal.

The angle $\theta$ immediately after the collision is determined by the ratio of the velocity components.

$$\tan\theta = \frac{v_y}{v_x}$$ $$\tan\theta = 1$$ $$\theta = 45^\circ = \frac{\pi}{4}$$

Answer: $\dfrac{\pi}{4}$ (or $45^\circ$)