Question

You’re in an airplane that flies horizontally with speed 1000 km/h (280 m/s) when an engine falls off. Ignore air resistance and assume it takes 30 s for the engine to hit the ground. (a) Show that the airplane is 4.5 km high. (b) Show that the horizontal distance that the aircraft engine moves during its fall is 8400 m. (c) If the airplane somehow continues to fly as if nothing had happened, where is the engine relative to the air- plane at the moment the engine hits the ground

Answer 1

(a)

For this part of the problem, we can ignore the horizontal motion of the engine and consider only the vertical motion.

The vertical position of the engine at time t is given by

$y(t) = h - \frac{1}{2}gt^2$

where

h is the initial altitude of the airplane

g = 9.81 m/s^2 is the acceleration due to gravity

t is the time

Since the engine takes 30 seconds to hit the ground, t = 30 s when y(t) = 0 (the ground). Substituting into the equation, we find h:

$0 = h - \frac{1}{2}gt^2\\h= \frac{1}{2}gt^2 = \frac{1}{2}(9.81 m/s^2)(30 s)^2=4,415 m \sim 4.5 km$

(b)

For this part of the problem, we can ignore the vertical motion and consider the horizontal motion only.

Since the engine travels at constant speed along the horizontal direction:

v = 280 m/s

its horizontal position after time t is given by

$x(t) = v t$

If we substitute

t = 30 s

which is the total duration of the fall, we can find the horizontal distance covered by the airplane during this time:

$x(t) = (280 m/s)(30 s)=8,400 m$

(c) The engine will be exactly 4.5 km under the plane

Here we have:

- the airplane is moving horizontally, at 4.5 km of altitude, at constant velocity of 280 m/s

- The engine moves both horizontally, also with a horizontal velocity of 280 m/s, and vertically, with acceleration 9.81 m/s^2 towards the ground

Both the plane and the engine moves with same horizontal velocity, so they cover the same horizontal distance (8400 m) during the 30 seconds. The only difference is that the engine falls down approx. 4.5 km, so the engine will be 4.5 km under the plane, when it hits the ground.

Answer 2

The airplane is 4.5 km high and the horizontal distance that the aircraft engine moves during its fall is 8400 m.

Newton Law of motion

To find the height if the plane from ground, we use:

H = ut + (1/2)gt²

Where u = initial velocity = 0 m/s, t = time = 30 s, g = acceleration due to gravity = 10 m/s²

H = 0(30) + (1/2)(10)(30)²

H = 4500 m = 4.5 km

The horizontal distance (R) is given by:

R = ut = 280(30) = 8400 m

Let d represent the distance from the engine to the air plane at the ground, hence:

d² = 8400² + 4500²

d = 9529 m

The airplane is 4.5 km high and the horizontal distance that the aircraft engine moves during its fall is 8400 m.

Find out more on Newton Law of motion at: https://brainly.com/question/25842103