The speed of a projectile when it reaches its maximum height is 0.42 times its speed when it is at half its maximum height. What is the initial projection angle of the projectile?

Answer 1

the max height is (v sinθ)^2/19.6
the speed at time t is √(Vx^2+Vy^2)
= √((v cosθ)^2 + (v sinθ - 9.8t)^2)

The speed at max height is just Vx = v cosθ

So, find t when h(t) is half the max height. That is, when

vsinθ t - 4.9t^2 = (v sinθ)^2/39.2

Then plug that into the speed formula.

Then, solve for θ when that speed is .42*vcosθ

Answer 2

65.17