A projectile is fired with speed u from a point at height h above a horizontal plane. The angle of projection θ may be varied.
If $\theta_{m}$ gives the maximum horizontal range, then:
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A projectile is fired with speed u from a point at height h above a horizontal plane. The angle of projection θ may be varied. If θ m 	heta_{m} θ m ​ gives the maximum horizontal range, then:

G1 Physics · Accepted Answer

For a trajectory from height h, R = (u cosθ/g) [u sinθ + √(u²sin²θ+2gh)]. A convenient geometric/resultant-velocity optimization gives the maximum-range condition tanθ_m = u/√(u²+2gh). For h=0 this reduces to tanθ_m=1, so θ_m=45°, as expected.

A projectile is fired with speed u from a point at height h above a horizontal plane. The angle of projection θ may be varied. If θ m \theta_{m} θ m ​ gives the maximum horizontal range, then:

A projectile is fired with speed u from a point at height h above a horizontal plane. The angle of projection θ may be varied. If θ m \theta_{m} θ m gives the maximum horizontal range, then: