The horizontal component of the projectile's initial velocity is Vx = vCosπ
The vertical component of the projectile's initial velocity is Vy = vSinπ
At the instant the projectile achieves its maximum height above ground level, it is displaced X = R/2 = (V²Sin2π)/2g
Assuming the projectile is projected at an angle π with the horizontal...
Therefore, resolving the projectile motion in the horizontal and vertical axis results in:
The horizontal component of the projectile's initial velocity is Vx = vCosπ
The vertical component of the projectile's initial velocity is Vy = vSinπ
Where v is the initial velocity of projection.
The time required to return back to its plane of projection ,T = (2vSinπ)/g
Therefore, the range, R of the projectile,
R = Vy × (2vSinπ)/g
R = vCosπ × (2vSinπ)/g
Recall, 2sinπcosπ = Sin2π
Therefore, R = v²sin2π/g.
Therefore, the horizontal displacement at Maximum height , X = R/2 = (v²Sin2π)/2g.
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