Question

A feather is dropped onto the surface of the moon. How far will the feather have fallen if it reaches in 9.00 seconds? Given: g on moon = -1.6 meters/second squared

Answer 1

The distance traveled while accelerating from rest is

                           D  =  1/2 a t²    .

For this problem, we shall totally ignore air resistance.
We do so completely without any reservation or guilt,
because we know that there is no air on the moon.

                     D  =  (1/2) · (1.6 m/s²) · (9 sec)²

                         =       (0.8 m/s²) · (81 s²)

                         =        (0.8 · 81) m

                         =            64.8 meters  .

(That's about 213 feet !  The astronaut must have dropped the feather
from his spacecraft while he was aloft ... either just before touchdown
or just after liftoff.)

Answer 2

Using equation of motion:

H = ut + (1/2)gt²,       

Where u = initial velocity = 0,  g = 1.6 m/s² (positive for falling downward),

t = time = 9 s

H = ut + (1/2)gt²

H = 0*9 + (1/2)*1.6*9²  = 64.8

So it would have fallen through 64.8 m.