Question

The speed of a projectile when it reaches its
maximum height is 1/5 the speed when the projectile is at 1/5
its maximum height.
What is the initial projection angle?

Answer 1

vertical problem first, initial speed = Vi = S sin theta
what is the speed 1/5 of the way up
starting speed up = Vi
starting energy = 1/2 m Vi^2 + m g*0
energy at top = m g H = 1/2 m Vi^2
energy at 1/5 = m g H/5 + (1/2) m v^2 = 1/2 m Vi^2
so
(1/2) m Vi^2- (1/2) m v^2 = m g H/5
5 v^2 = 5 Vi^2 - 2 g H
but 2 g H = Vi^2
5 v^2 =4 Vi^2
v = Vi sqrt(4/5)

now speed at top = S cos theta
speed at 1/5 = sqrt [ (S cos theta)^2 + (S sin theta * sqrt4/5)^2]
so
1/5 = S cos theta / sqrt [ (S cos theta)^2 + (S sin theta * sqrt4/5)^2]
5 cos theta = sqrt [ ( cos theta)^2 + ( sin theta * sqrt4/5)^2]
5 cos theta = sqrt (cos^2 theta + 0.8 sin^2 theta )
24 cos^2 theta = 0.8 sin^2 theta
tan^2 theta = 30
tan theta = 5.48
theta = 79.6 deg

Answer 2

I did that fast, check the math!!!