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/s^2 downward, how long does it take the feather to hit the Moon's surface?
h=1/2 a t^2 = 1/2 (1.62)t^2=1.1
solve for t
To determine how long it takes for the feather to hit the surface of the moon, we can use the kinematic equation that relates time, initial velocity, acceleration, and distance. The equation is as follows:
s = ut + (1/2)at^2
where:
s is the distance (in this case, the distance the feather drops),
u is the initial velocity (which is 0 because the feather is dropped at rest),
a is the acceleration due to gravity on the moon (1.62 m/s^2), and
t is the time it takes for the feather to hit the surface.
Plugging in the given values, the equation becomes:
1.1 = 0*t + (1/2)*(1.62)*t^2
Simplifying further:
1.1 = (0.81)*t^2
To solve for t, we need to rearrange the equation by dividing both sides by 0.81:
t^2 = 1.1 / 0.81
t^2 = 1.358
Taking the square root of both sides, we get:
t = √1.358
t ≈ 1.164 seconds
Therefore, it takes approximately 1.164 seconds for the feather to hit the surface of the moon.