Answer:
v₀ₓ = 15 m / s, [tex]v_{oy}[/tex] = 5.2 m / s
v = 15.87 m / s , θ = 19.1
Explanation:
This is a projectile launch problem. The horizontal speed that is constant throughout the entire path is worth 15 m / s, instead the vertical speed changes in value due to the acceleration of gravity, let's look for the initial vertical speed
Vy² =[tex]v_{oy}[/tex]² - 2 g y
[tex]v_{oy}[/tex]² = [tex]v_{y}[/tex]² + 2 g y
[tex]v_{oy}[/tex] = √ ([tex]v_{y}[/tex]² + 2 gy
Let's calculate
[tex]v_{oy}[/tex] = √ (1.25² + 2 9.8 1.3)
[tex]v_{oy}[/tex] = √ (27.04)
[tex]v_{oy}[/tex] = 5.2 m / s
The initial speed can be calculated by the initial speed
v = √ v₀ₓ² + [tex]v_{oy}[/tex]²
v = RA (15² + 5.2²)
v = 15.87 m / s
We look for the angle with trigonometry
tan θ = voy / vox
θ = tan⁻¹ I'm going / vox
θ = tan⁻¹ 5.2 / 15
θ = 19.1
The answer is
v₀ₓ = 15 m / s
[tex]v_{oy}[/tex] = 5.2 m / s