A projectile is launched up and to the right over flat, level ground. Its range is equal to half of its maximum elevation above the ground. What was the angle between its initial velocity and the ground? Ignore air resistance.

Question

° above the horizonontal

Answer 1

For projectile motion,

Range = R = (u^2)(sin2θ)/g

Maximum height = H = (u^2)(sinθ)^2/(2g)

As per the question, 2R = H

=> 2(u^2)(sin2θ)/g = (u^2)(sinθ)^2/(2g)
=> 4sin2θ = (sinθ)^2
=> 8sinθcosθ = (sinθ)^2
=> 8cosθ = sinθ
=> tanθ = (8)
=> θ = arctan(8)