Respuesta :
Answer:
a) Â Â v = 520.89 m / s , b) Â Â x = 11,186.89 m
Explanation:
This is a projectile launch exercise. In this case the plane travels horizontally, so its initial vertical speed is zero, let's calculate the time it takes to reach the ground
     y = y₀ + v₀t - ½ gt²
     0 = y₀ - ½ g t²
     t = √ (2y₀ / g)
     t = √ (2 11000 / 9.8)
     t = 47.38 s
the speed when reaching the ground is
     vₓ = 850 km / h
let's reduce it to SI units
     vₓ = 850 km / h (1000m / 1 km) (1 h / 3600s)
      vₓ = 236.11 m / s
vertical speed is
      [tex]v_{y}[/tex] = go - gt
      v_{y} = -9.8 47.38
      v_{y} = - 464.3 m / s
we use the Pythagorean theorem to find total velocity
      v = √ (vₓ² + [tex]v_{y}[/tex]²)
      v = √ (236.11² +464.3²)
      v = 520.89 m / s
this speed is not too high, but the reality is that the friction with the air will decrease this significant speed,
b) the range from where the plane will fall
       x = vx t
       x = 236.11 47.38
       x = 11,186.89 m
c) the plane will fall much smaller backfire due to air friction