A ground-to-ground projectile has horizontal range R and maximum height H. Its projection speed u satisfies:
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A ground-to-ground projectile has horizontal range R and maximum height H. Its projection speed u satisfies:

G1 Physics · Accepted Answer

For a projectile, R = u²sin2θ/g, H = u²sin²θ/(2g). Also R = 4H cotθ ⇒ tanθ = 4H/R. Using H = u²sin²θ/(2g), and sin²θ = tan²θ/(1+tan²θ), we get u² = 2gH(1+tan²θ)/tan²θ = 2gH[1 + R²/(16H²)] = g(R²+16H²)/(8H).