a projectile is fired with an initial speed of 53 m/s. Find the angle of projection such that the maximum height of the projectile is equal to its horizontal range

Answer 1

range is v^2 sin(2θ)/g
height is v^2 sin^2(θ)/2g

so,

sin(2θ) = sin^2(θ)/2
2sinθcosθ = sin^2(θ)/2
4cosθ = sinθ
16cos^2(θ) = sin^2(θ) = 1 - cos^2(θ)
17cos^2(θ) = 1
cosθ = 1/√17

Answer 2

0.34 & 0.125