Question

derivation of maximum height of projectile​

Answer 1

Answer:

if α = 45°, then the equation may be written as: hmax = h + V₀² / (4 * g) and in that case, the range is maximal if launching from the ground (h = 0). if α = 0°, then vertical velocity is equal to 0 (Vy = 0), and that's the case of horizontal projectile motion

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