Respuesta :
Answer:
v₀ = 14.6 m / s
Explanation:
This is a projectile launch exercise, in this case we must stop the wall, we can assume that at this point the ball is at its maximum height
as the height is maximum its vertical speed is zero
v_{y} = [tex]v_{oy}[/tex] - g t
y = v_{oy} t - ½ g t²
the horizontal distance is
x = v₀ₓ t
we use trigonometry to find the components of the velocity
sin θ = v_{oy} / v₀
cos θ = v₀ₓ / v₀
v₀ₓ = v₀ cos θ
v_{oy} = v_{o} sin θ
we substitute the values
0 = v₀ sin θ - g t
10 = v₀ sin θ t - ½ g t²
6 = v₀ cos θ t
we have three equations and three unknowns by which the system can be solved
let's use the first and second equations,
10 = (gt) t - ½ g t²
10 = g t² - ½ g t²
10 / 9.8 = t² (1- ½)
1.02 = ½ t²
t = √ 2.04
t = 1,429 s
this is the time to hit the wall
now let's use the first and third equations, we substitute the data
v₀ sin θ = g t
v₀ cos θ = x / t
we divide and simplify
tan θ = g t² / x
θ = tan⁻¹ g t² / x
θ = tan⁻¹ (9.8 1.429²/6)
θ = 73.3°
we substitute in the third equation
x = v₀ cos 73.3 t
v₀ = x / t cos 73.3
v₀ = 6 / (1.429 cos 73.3)
v₀ = 14.6 m / s
