A feather is dropped on the moon from a height of 1.40meters.The acceleration of gravity on the moon is 1.67m/s^2.Determine the time for the feather to fall to the surface of the moon.
h=1/2 a t^2
t= sqrt(2h/a)
To determine the time it takes for the feather to fall to the surface of the moon, we can use the equation of motion for a freely falling object:
s = ut + (1/2)at^2
where:
s = distance traveled (1.40 meters)
u = initial velocity (0 m/s, as the feather is dropped and has no initial velocity)
a = acceleration due to gravity (-1.67 m/s^2, since the feather is falling towards the moon)
t = time
Rearranging the equation, we have:
1.40 = (1/2)(-1.67)t^2
To solve for t, we can isolate t by multiplying both sides of the equation by 2 to get rid of the fraction:
2 × 1.40 = -1.67t^2
2.80 = -1.67t^2
Next, divide both sides of the equation by -1.67 to solve for t^2:
t^2 = 2.80 / -1.67
t^2 ≈ -1.68
Since time cannot be negative, we can conclude that there is no real solution for t in this case.