a projectile is fired so that when it reaches the maximum height h it has speed vtop when the projectile is at h/2 the speed is 2vtop at what angle is the projectile fired

Explanation:

Let the velocity of firing be u at angle θ

At maximum height velocity will be equal to horizontal component of initial velocity or vcosθ

So , vtop = v cosθ

At height h/2

vertical component of velocity v₂

v₂² = (usinθ)² - 2 g . h/2

v₂² = u²sin²θ - gh

horizontal component of velocity at height h/2 = u cosθ

velocity at height h / 2

= √ ( u²sin²θ - gh + u² cos²θ)

Given

√ ( u²sin²θ - gh + u² cos²θ) = 2 vtop

u²sin²θ - gh + u² cos²θ = 4 v²top = 4 u² cos²θ

u²sin²θ - gh =  3 u² cos²θ

At height h , vertical component of velocity is zero

0 = u²sin²θ - 2gh

gh = u²sin²θ / 2

u²sin²θ - u²sin²θ / 2  =  3 u² cos²θ

u²sin²θ / 2  = 3 u² cos²θ

Tan²θ = 6

Tanθ = 2.45

θ = 68⁰ .  

it was given as a lesson and guide for prayer.

the right answer is 0.3 meters per second