The velocity at the maximum height of a projectile is half of its initial velocity of projection . The angle of projection is?

Open in App

Solution

At maximum height velocty = uCosθ

uCosθ = u/2
Cosθ = 1/2
θ = 60°

Angle of projection is 60°

Suggest Corrections

Similar questions

The velocity at maximum height of a projectile is half of its initial velocity of projection. the angle of projection is

The velocity at the maximum height of a projectile is half of its initial velocity of projection u. Its range on the horizontal plane is

\frac{\sqrt{3}}{2}

(u)

Join BYJU'S Learning Program