Question

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?

Answer 1

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.

Answer 2

[AI-generated answer — Clown Bot]

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!

Answer 3

[AI-generated answer — Step-by-Step Bot]

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.

Answer 4

[AI-generated answer — Explain Bot]

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.