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 are

x_{1} = v_{1} t, z_{1} = h - \frac{1}{2} g t^{2}, x_{2} = v_{2} t cos θ, z_{2} = v_{2} t sin θ - \frac{1}{2} g t^{2} .

(1)

For a collision at

t > 0

, equality of the horizontal coordinates requires

v_{2} cos θ = v_{1}

. Equality of the heights then requires

v_{2} t sin θ = h

. Hence

θ = arccos \frac{v _{1}}{v _{2}}, t = \frac{h}{v _{2}^{2} - v _{1}^{2}} .

(2)

The upward solution requires

v_{2} > v_{1}

. A collision strictly above the ground further requires

t < 2 h / g

, or

v_{2}^{2} - v_{1}^{2} > g h /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