 Solution

Take horizontal motion towards the second projectile as positive. In projectile motion, both bodies have the same downward gravitational acceleration, so it cancels from their relative motion. Their coordinates areFor a collision at $t>0$, equality of the horizontal coordinates requires $v_2\cos \theta =v_1$. Equality of the heights then requires $v_{2}t\sin \theta =h$. HenceThe upward solution requires $v_2>v_1$. A collision strictly above the ground further requires $t<\sqrt{2h/g}$, or $v_2^2-v_1^2>gh/2$. The horizontal target distance does not independently affect the interception geometry. If the first shot is also required to hit that ground target, its parameters satisfy $l=v_1\sqrt{2h/g}$.