Suppose an astronaut drops a feather from 1.1 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?
distance = 1/2 a t^2 ... 1.1 = 1/2 * 1.62 * t^2
solve for t
To find the time it takes for the feather to hit the Moon's surface, we can use the kinematic equation for displacement:
d = vi * t + (1/2) * a * t^2,
where:
- d is the displacement (in this case, the distance the feather falls)
- vi is the initial velocity of the feather (0 m/s, since it's dropped from rest)
- a is the acceleration due to gravity on the Moon (-1.62 m/s^2, because it is downward)
- t is the time it takes for the feather to hit the Moon's surface
Since the feather is dropped from rest, its initial velocity (vi) is 0 m/s. Therefore, the equation becomes:
d = (1/2) * a * t^2.
We can rearrange the equation to solve for t:
t^2 = (2 * d) / a.
Substituting the given values:
d = 1.1 m,
a = -1.62 m/s^2,
we get:
t^2 = (2 * 1.1 m) / (-1.62 m/s^2).
Calculating the right side of the equation:
t^2 = -2.2 m / (-1.62 m/s^2).
Dividing both sides by -1 (to eliminate the negative sign):
t^2 = (2.2 m) / (1.62 m/s^2).
Now, taking the square root of both sides:
t ≈ √(2.2 m / 1.62 m/s^2).
Evaluating this expression, we find:
t ≈ √1.358 m/s^2.
Finally, calculating the square root:
t ≈ 1.165 s.
Therefore, it takes approximately 1.165 seconds for the feather to hit the Moon's surface.