Final answer:
The launch angle θ₀ that corresponds to a projectile's launch speed being twice its speed at maximum height is 60°.
Step-by-step explanation:
If a projectile's launch speed is twice its speed at maximum height, we are dealing with the conservation of energy where all potential energy at max height was converted from kinetic energy at launch less any speed lost to air resistance, which we are ignoring in this ideal scenario. Since the only force acting on the projectile is gravity, the speed at the maximum height will be entirely horizontal. This is because at the maximum height, the vertical component of the velocity is zero.
At launch, the projectile has both vertical and horizontal components of velocity. Given that the launch speed is twice the speed at maximum height, which is entirely horizontal, we can set up the ratio of the horizontal component (vx) to the total initial launch speed (v0), where vx = v0 cos(θ0). Since at maximum height vx = 1/2 v0, we have 1/2 v0 = v0 cos(θ0). Solving for the angle, we find that cos(θ0) = 1/2, which corresponds to θ0 = 60°.