Suppose an astronaut drops a feather from 1.6 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?
h = 0.5*g*t^2 = 1.8.
0.5*1.62*t^2 = 1.8,
t =
To find the time it takes for the feather to hit the Moon's surface, we can use the equation of motion:
s = ut + (1/2)at^2
Where:
s = distance
u = initial velocity
t = time
a = acceleration due to gravity
In this case, the feather is dropped from rest (u = 0), and the distance (s) is given as 1.6 m. The acceleration due to gravity (a) on the Moon is given as 1.62 m/s^2.
Plugging in the values, the equation becomes:
1.6 = 0*t + (1/2)(1.62)*t^2
Simplifying the equation:
1.6 = 0 + 0.81*t^2
Dividing both sides of the equation by 0.81:
1.6/0.81 = t^2
Solving for t:
t^2 = 1.9753
Taking the square root of both sides:
t ≈ 1.405 seconds
Therefore, it takes approximately 1.405 seconds for the feather to hit the Moon's surface.