Question

If velocity of an oblique projectile at maximum
height is half of the initial velocity u, then time
of flight of the projectile is​

Answer 1

Answer:

Explanation:

At maximum height

velocity = uCosθ

u/2 = uCosθ

θ = 60°

R = u² Sin(2θ) / g

= u² Sin(2 × 60°) / g

= u²√3 / (2g)

Range on horizontal plane is u²√3 / (2g)2