a 15000 kg airplane descends a vertical distance of 10.0 km over a horizontal distance of 80.0 km. What is the airplane's loss of potential energy.
(I'm confused about the height)
Ep=mgh
m= 15000 kg
g= 9.81 m/s^2
h=??
10,000 meters
the horizontal distance is irrelevant.
Thanks!
You are welcome. Remember potential is always a difference (In height, voltage, whatever).
To find the height (h) in the formula Ep = mgh, we need to relate the horizontal distance (80.0 km) to the vertical distance (10.0 km).
Since the airplane is descending vertically, the vertical distance is the change in height. We can treat the horizontal distance as the hypotenuse of a right triangle, where the vertical distance is the opposite side and the height is the adjacent side.
Using the Pythagorean theorem, we can calculate the height (h):
h^2 = hypotenuse^2 - opposite^2
h^2 = (80.0 km)^2 - (10.0 km)^2
h^2 = 6400 km^2 - 100 km^2
h^2 = 6300 km^2
To find h, we take the square root of both sides:
h = sqrt(6300 km^2)
h ≈ 79.37 km
Now that we have the height (h ≈ 79.37 km), we can calculate the potential energy change using the given mass (m = 15000 kg) and the acceleration due to gravity (g = 9.81 m/s^2).
Ep = mgh
Ep = (15000 kg)(9.81 m/s^2)(79.37 km)
Ep ≈ 117218850 J
Therefore, the airplane's loss of potential energy is approximately 117,218,850 joules.