Suppose an astronaut drops a feather from 1.4 m above the surface of the Moon. If the acceleration due to gravity on the Moon is 1.62 m/s2 downward, how long does it take the feather to hit the Moon's surface?

Solve the equation

(1/2)(1.62) t^2 = 1.4
That will give you the time, t

1.34

To find out how long it takes for the feather to hit the Moon's surface, we can use the equation of motion:

h = ut + (1/2) * g * t^2

Where:
h = initial height of the feather (1.4 m)
u = initial velocity of the feather (which is 0 m/s since it is dropped)
g = acceleration due to gravity on the Moon (-1.62 m/s^2)
t = time taken for the feather to hit the surface

Since the feather is dropped, the initial velocity, u, is 0 m/s. Thus, our equation becomes:

h = (1/2) * g * t^2

Substituting the given values, we have:

1.4 = (1/2) * (-1.62) * t^2

Simplifying the equation gives:

1.4 = -0.81 * t^2

Dividing both sides by -0.81:

t^2 = -1.4 / -0.81

t^2 = 1.7284

Taking the square root of both sides:

t = √1.7284

t ≈ 1.314 seconds

Therefore, it takes approximately 1.314 seconds for the feather to hit the Moon's surface.