an astronaut drops a feather from 1.2 m above the surface of the moon. if the acceleration of the gravity on the moon is 1.62 m/s down , how long does it take the feather tho hit the surface

To find the time it takes for the feather to hit the surface of the moon, we can use the kinematic equation:

\(d = \frac{1}{2}gt^2\)

where:
- d is the distance (1.2 m in this case),
- g is the acceleration due to gravity on the moon (-1.62 m/s² since it acts downward),
- t is the time.

First, we rearrange the equation to solve for time (t):

\(t^2 = \frac{2d}{g}\)

\(t = \sqrt{\frac{2d}{g}}\)

Now, we can substitute the given values into the equation to calculate the time:

\(t = \sqrt{\frac{2 \times 1.2}{-1.62}}\)

\(t = \sqrt{\frac{2.4}{-1.62}}\)

\(t = \sqrt{-1.48}\)

Since the square root of a negative number is not defined in real numbers, this result tells us that the feather will not hit the surface of the moon. This is because the feather experiences a very small gravitational force compared to its air resistance, causing it to float and take a longer time to reach the ground.

retg