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?

Well, since the feather is on the Moon, it's already in an out-of-this-world situation! *ba dum tss* Now, let's calculate the time it takes for our feather to gracefully reach the Moon's surface.

Using the equation of motion: s = ut + (1/2)at^2, where s is the distance, u is the initial velocity, a is the acceleration, and t is the time.

Given:
s = 1.4 m (distance)
u = 0 m/s (initial velocity)
a = 1.62 m/s^2 (acceleration)

Plugging in these values, we get:
1.4 = 0 x t + (1/2) x 1.62 x t^2

Arranging the equation, we have:
0.81t^2 = 1.4

To find t, we can take the square root of both sides:
t = √(1.4 / 0.81)

*tapping calculator keys*

And the answer is approximately 1.17 seconds! So, the feather takes about 1.17 seconds to touch down on the Moon's surface. Feel free to imagine it gracefully gliding down like a moon-bound ballerina!

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

d = vit + 0.5at^2

Where:
d = distance (1.4 m)
vi = initial velocity (0 m/s as the feather is dropped)
a = acceleration (-1.62 m/s^2, negative because it is downward)
t = time (unknown)

Plugging in the given values:

1.4 = 0 * t + 0.5 * (-1.62) * t^2

Simplifying the equation:

1.4 = -0.81 * t^2

Dividing both sides of the equation 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 surface of the Moon.

To determine the time it takes for the feather to hit the Moon's surface, we can use the kinematic equation for free fall:

\[ d = v_0 t + \frac{1}{2} a t^2 \]

where:
- \( d \) is the distance (1.4 m in this case)
- \( v_0 \) is the initial velocity (0 m/s since the feather is dropped)
- \( a \) is the acceleration due to gravity on the Moon (-1.62 m/s\(^2\) since it is downward)
- \( t \) is the time we want to calculate

Rearranging this equation, we can solve for \( t \):

\[ t = \sqrt{\frac{2d}{-a}} \]

Substituting the given values, we have:

\[ t = \sqrt{\frac{2 \times 1.4}{-1.62}} \]

Now, let's calculate the answer:

\[ t = \sqrt{\frac{2 \times 1.4}{-1.62}} \approx 0.895 \, \text{s} \]

Therefore, it takes approximately 0.895 seconds for the feather to hit the Moon's surface.

Ignore air resistance, which is justifiable in an insignificant atmosphere.

S=ut+(1/2)at²
u=initial velocity=0
a=acceleration du to gravity=-1.62 m/s/s
S=displacement = -1.4 m (downwards)
t=time in seconds

-1.4 = 0*t + (1/2)(-1.62)t²
t=sqrt(2*1.4/1.62)
=1.31 s.