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?
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To find the time it takes for the feather to hit the Moon's surface, we can use the equation of motion that relates distance, initial velocity, acceleration, and time.
The equation we can use is:
d = vit + (1/2)at^2
Where:
d = distance
vi = initial velocity
a = acceleration
t = time
In this case, the feather is dropped, so the initial velocity (vi) is 0.
The acceleration due to gravity on the moon is given as 1.62 m/s^2 downward. Since the feather is dropping, we take this acceleration as a positive value.
The distance the feather needs to fall is 1.1 m.
Plugging in these values into the equation, we get:
1.1 = 0 + (1/2)(1.62)(t^2)
Simplifying further:
1.1 = 0.81t^2
Dividing both sides by 0.81:
t^2 = 1.358
Taking the square root of both sides:
t ≈ √1.358
t ≈ 1.165 seconds
Therefore, it takes approximately 1.165 seconds for the feather to hit the Moon's surface.