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A jet aircraft is travelling at 850 km/hr at an altitude of 11 km – when an engine falls off! a. Neglecting air resistance, determine the speed of the engine when it hits the ground. Is this realistic? Discuss how assumptions made in the calculations might not reflect the reality of the engine's impact speed. What value might the impact speed actually be and why? b. Determine the horizontal range of the engine (measured from where the jet was located when the engine fell off). c. Reality – Would the engine land at this point in reality? Why or why not? Under what conditions would your solution be valid? Where (roughly) would it actually land?

1 Answer

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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


v_(y) = 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 +
v_(y)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

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