The velocity at the maximum height is root 3/2 times of its initial velocity of projection u. It's horizontall range is
Answers
Answered by
2
Velocity at maximum height = uCosθ
According to question
uCosθ = (√3 / 2) × u
Divide both sides by u
Cosθ = √3 / 2
Cosθ = Cos30
θ = 30°
Angle of projection is 30°
Horizontal range = u² sin(2θ)/g
= u² sin(2 × 30) / g
= u² / (2g)
It’s Horizontal range is u² / (2g)
According to question
uCosθ = (√3 / 2) × u
Divide both sides by u
Cosθ = √3 / 2
Cosθ = Cos30
θ = 30°
Angle of projection is 30°
Horizontal range = u² sin(2θ)/g
= u² sin(2 × 30) / g
= u² / (2g)
It’s Horizontal range is u² / (2g)
Similar questions