Aprojectile is launched from the earth’s surface at initial speed v0 at angle θ0 with the horizontal. when the projectile is at its maximum height h, it has half the speed it had when it was at half its maximum height h 2 . at what angle was the projectile launched.

θ₀ = 67.79°

Explanation:

Given info

we know that

Ymax = v₀y² / (2g)

v = v₀x  (when Y = Ymax)

when the projectile was at half its maximum height (Y')

v' = 2v = 2*v₀x

we can use the equations

(v')²= v'x² + v'y²   ⇒  (2*v₀x)² = (v₀x)² + v'y²   ⇒   v'y² = 3*(v₀x)²   (I)

if we know that

v'y² = v₀y² - 2*g*(y')  

y' = Ymax /2 = (v₀y² / (2g)) / 2 = v₀y² / (4g)

then

v'y² = v₀y² - 2*g*(v₀y² / (4g)) = v₀y² - (v₀y² / 2) = v₀y² / 2

⇒ v'y² = v₀y² / 2       (II)

we can say that (I) = (II)

3*(v₀x)² = v₀y² / 2   ⇒   v₀y = √6*v₀x

Finally we apply

tan θ₀ = v₀y / v₀x

⇒  tan θ₀ = √6*v₀x / v₀x = √6

⇒  θ₀ = tan⁻¹ (√6) = 67.79°

i think they are all correct

explanation:

they seem good

not too sure thinking all accelerations are pointed down