An airplane takes 4 hours to travel a distance of 3600 miles with the wind. the return trip takes 5 hours against the wind. find the speed of the plane in still air and the speed of the wind.

wind speed = 90 mi/h

aircraft speed = 810 mi/h

Explanation:

Let the speed of the airplane be u and the wind speed be w.

The distance d traveled is given by: d = speed * time

1. trip:

d = 3600 miles, time t = 4 h, speed = u + w

3600 = (u + w) * 4

2. trip:

d = 3600 miles, time t = 5 h, speed = u - w

3600 = (u -w) * 5

You have two equations with two unknown. Solve for u and w.

For example:

(u + w) * 4 = (u - w) * 5

4u + 4w = 5u - 5w

u = 9w

Plugging this result in one of the equations gives:

3600 = (9w + w) * 4

3600 = 10w * 4

w = 90 mi/h

u = 810 mi/h

answer: point c: b. 1/2 potential, 1/2 kinetic, point a: 3/4 of its potential energy, point d or b: 3/4 potential, 1/4 kinetic

explanation: i cant explain. it's difficult to explain.

the device is called “ rocket engine”