Respuesta :
Answer:
v_{f} = 74 m/s, F = 230 N
Explanation:
We can work on this exercise using the relationship between momentum and moment
    I = ∫ F dt = Δp
bold indicates vectors
we can write this equations in its components
X axis
    Fₓ t = m ( -v_{xo})
Y axis Â
    t = m (v_{yf} - v_{yo})
in this case with the ball it travels horizontally v_{yo} = 0
Let's use trigonometry to write the final velocities and the force
    sin 30 = v_{yf} / vf
    cos 30 = v_{xf} / vf
    v_{yf} = vf sin 30
    v_{xf} = vf cos 30
     sin40 = F_{y} / F
     F_{y} = F sin 40
     cos 40 = Fₓ / F
     Fₓ = F cos 40
let's substitute
   F cos 40 t = m ( cos 30 - vₓ₀)
   F sin 40 t = m (v_{f} sin 30-0)
we have two equations and two unknowns, so the system can be solved
    F cos 40 0.1 = 0.4 (v_{f} cos 30 - 20)
    F sin 40 0.1 = 0.4 v_{f} sin 30
we clear fen the second equation and subtitles in the first
     F = 4 sin30 /sin40   v_{f}
     F = 3.111 v_{f}
    (3,111 v_{f}) cos 40 = 4 v_{f} cos 30 - 80
    v_{f} (3,111 cos 40 -4 cos30) = - 80
    v_{f} (- 1.0812) = - 80
    v_{f} = 73.99
    v_{f} = 74 m/s
now we can calculate the force
     F = 3.111 73.99
     F = 230 N
