Answer:
a) v = 520.89 m / s , b) x = 11,186.89 m
Step-by-step 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
= 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 +
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