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?
Solve the equation
(1/2)(1.62) t^2 = 1.4
That will give you the time, t
1.34
To find out how long it takes for the feather to hit the Moon's surface, we can use the equation of motion:
h = ut + (1/2) * g * t^2
Where:
h = initial height of the feather (1.4 m)
u = initial velocity of the feather (which is 0 m/s since it is dropped)
g = acceleration due to gravity on the Moon (-1.62 m/s^2)
t = time taken for the feather to hit the surface
Since the feather is dropped, the initial velocity, u, is 0 m/s. Thus, our equation becomes:
h = (1/2) * g * t^2
Substituting the given values, we have:
1.4 = (1/2) * (-1.62) * t^2
Simplifying the equation gives:
1.4 = -0.81 * t^2
Dividing both sides 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 Moon's surface.