Physics
posted by Amanda and Travis on .
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

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 
0.34 & 0.125