BullsEye

$x$ -axis movement

$\boxed{V_0\cos {\alpha t}=d}$

$y$ -axis movement

$h-x$ is the height at the maximum point of the projection

From body $A$ until $h-x$ height

$h-x=h-4.9t^2$

$-x=-4.9t^2$

From body $B$ until $h-x$ height

$h-x=V_0\sin {\alpha t}-4.9t^2$

Then

$h-x=V_0\sin {\alpha t}-x$

$\boxed{h=V_0\sin {\alpha} t}$

Finally

$\frac{V_0\sin {\alpha} t}{V_0\cos {\alpha} t}=\frac{h}{d}\Rightarrow \tan{\alpha}=\frac{h}{d} \therefore \boxed{\alpha=\tan^{-1} \left( \frac{h}{d}\right)}$

In this question we can simply get to the result in one line if we apply relative motion between two projectiles..!!

For the particles to hit each other, initial velocity of B relative to A must point towards A. If at initial situation this condition is met then the velocity of B wrt A will always be pointing towards A during the motion.

Basically both of them has the same accelerations and their relative acceleration is zero..therefore in frame of A, B will appear to be moving in a straight line. If this straight line is pointing towards A then B will eventually hit A.

So simply speaking, $\tan\alpha\ = \frac{h}{d}$