Question

If R and H represent horizontal range and
maximum height of the projectile, then the
angle of projection with the horizontal is​

Answer 1

Explanation:

H = [u^2 (sinθ)^2]/2g

R = (u^2)sin2θ/g

H/R = sinθ/2cosθ

tanθ = 4H/R

θ = tan(^-1)(4H/R)

Answer 2

Answer:

tan

$\tan( \alpha ) = 4hmax \div r$

Explanation:

if tan 45°=1

R = 4 Hmax