The aircraft taking off, rising to an altitude of 9 km, picks up speed of 720 km / h.

The aircraft taking off, rising to an altitude of 9 km, picks up speed of 720 km / h. Compare the kinetic and potential energies acquired by the aircraft: which of them is greater and how many times?

h = 9 kilometers = 9000 meters – the height to which the plane climbed;

g = 9.8 meters per second squared – gravitational acceleration;

v = 720 meters per second – the speed that the plane gained.

It is required to determine which of the energies (potential or kinetic) has increased more and by a factor of magnitude.

At the beginning, the plane was on the surface of the earth and did not move, which means that its potential and kinetic energy are equal to 0.

Let m be the mass of the aircraft. Then, we find its potential energy:

Epot = m * g * h = m * 9.8 * 9000 = 88200 * m Joules.

Let’s find its kinetic energy:

Ekin = m * v ^ 2/2 = m * 7202/2 = m * 518400/2 = 259200 * m Joules.

It is obvious that Ekin> Epot, since 259200 * m> 88200 * m.

Wherein:

Ekin / Epot = 259200 * m / (88200 * m) = 2.9 times.

Answer: the kinetic energy was 2.9 times greater.