the velocity of maximum height of projectile is half of its initial velocity of projection the angle of projection is
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Velocity at maximum height = uCosθ
According to question
uCosθ = u/2
Divide both sides by “u”
Cosθ = 1/2
Cosθ = Cos60
θ = 60°
∴ Angle of projection is 60°
According to question
uCosθ = u/2
Divide both sides by “u”
Cosθ = 1/2
Cosθ = Cos60
θ = 60°
∴ Angle of projection is 60°
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