Answer:
x = 27.3 m
Explanation:
This is a projectile launching exercise, let's start by looking for the time it takes for the rock to reach the height of the window.
Let's use trigonometry to find the velocities of the rock
      sin 40 = [tex]v_{oy}[/tex] / v
      cos 40 = v₀ₓ / v
      v_{oy}= v sin 40
      v₀ₓ = v cos 40
      v_{oy} = 30 sin 40 = 19.28 m / s
      v₀ₓ = v cos 40
      v₀ₓ = 30 cos 40 = 22.98 m / s
we look for the time
    [tex]v_{y}^2[/tex] = v_{oy}^2 - 2 g y
     v_{y}^2 = 19.28 2 - 2 9.8 16 = 371.71 - 313.6 = 58.118
    v_{y} = 7.623 m / s
we calculate the time
     v_{y} = v_{oy} - gt
     t = (v_{oy} - v_{y}) / g
     t = (19.28 -7.623) / 9.8
     t = 1,189 s
     Â
since the time is the same for both movements let's use this time to find the horizontal distance
      x = v₀ₓ t
      x = 22.98 1,189
      x = 27.3 m
