Compare the free fall times of bodies on the Earth and on the Moon, if the height from which the bodies

Compare the free fall times of bodies on the Earth and on the Moon, if the height from which the bodies fall is the same. Free fall acceleration on the Moon 1.6 m / s2

Given:

h1 = h2 = h – the height from which the bodies fall is the same;

g1 = 9.8 m / s2 is the acceleration of gravity on Earth;

g2 = 1.6 m / s2 – acceleration of gravity on the Moon.

It is required to determine t1 / t2 – the ratio of the free fall time of a body on the Earth and the Moon.

According to the condition of the problem, the falling of the bodies is free, that is, without the initial velocity.

Then, the time of the body falling on Earth will be equal to:

t1 = (2 * h1 / g1) ^ 0.5 = (2 * h / g1) ^ 0.5.

The time of the body falling on the moon will be equal to:

t2 = (2 * h2 / g2) ^ 0.5 = (2 * h / g2) ^ 0.5.

t1 / t2 = (2 * h / g1) ^ 0.5 / (2 * h / g2) ^ 0.5 = (g2 / g1) ^ 0.5 = (1.6 / 9.8) ^ 0, 5 = = 0.16 ^ 0.5 = 0.4. That is,

t1 = 0.4 * t2.

Answer: the time of the body falling on the Earth will be 0.4 times less than on the Moon.