Answer:
v = 47.85 m / s , θ = 64.7º
Explanation:
This is a missile throwing exercise.
Let's find the speed to reach the maximum height, at this point the vertical speed is zero
[tex]v_{y}^{2}[/tex] = v_{oy}^{2} - 2 g y
0 = v_{oy}^{2} - 2gy
v_{oy} = √2gy
let's calculate
v_{oy} = √ (2 9.8 21.3)
v_{oy} = 20.43 m / s
now we can calculate the time it takes to get to this point
vy = v_{oy} - g t
t = v_{oy} / g
t = 20.43 / 9.8
t = 2.08 s
in projectile launching, the time it takes for the body to rise is the same as the time it takes to go down, so the total launch time is
[tex]t_{v}[/tex] = 2 t
t_{v} = 2 2.08 = 4.16 s
let's use the horizontal throw ratio
x = v₀ₓ t_{v}
v₀ₓ = x / t_{v}
v₀ₓ = 180 / 4.16
v₀ₓ = 43.27 m / s
initial velocity is
v = √ (v₀ₓ² + v_{oy}^{2})
v = √ (20.43² + 43.27²)
v = 47.85 m / s
with an angle of
tan θ = I go / vox
θ = tan⁻¹ (43.27 / 20.43)
θ = 64.7º
