a feather is dropped on the moon from a height of 1.4 meters.the acceleration of the moon is 1.67 m/s determine the time for the feather to fall on the surface of
the moon
Assuming no wind:
Vo*t + 0.5g*t^2 = 1.4 m.
0 + 0.5*1.63*t^2 = 1.4
0.815t^2 = 1.4
t^2 = 1.72
t = 1.31 s.
To determine the time it takes for the feather to fall on the surface of the moon, we can use the kinematic equation:
s = ut + (1/2)at^2
where:
s = distance (1.4 meters)
u = initial velocity (0 m/s, as the feather is dropped)
a = acceleration (1.67 m/s^2)
t = time (unknown)
Rearranging the equation, we get:
1.4 = 0*t + (1/2)(1.67)t^2
1.4 = (0.835)t^2
Dividing both sides by 0.835, we have:
t^2 = 1.4/0.835
t^2 ≈ 1.6766
Taking the square root of both sides, we find:
t ≈ √1.6766 ≈ 1.296 seconds
Therefore, the time it takes for the feather to fall on the surface of the moon is approximately 1.296 seconds.
To determine the time it takes for the feather to fall on the surface of the moon, we can use the equation of motion:
s = ut + (1/2)at^2
Where:
s = distance (1.4 meters)
u = initial velocity (0 m/s since the feather is dropped)
a = acceleration (1.67 m/s^2 on the moon)
t = time
Since the feather is dropped, the initial velocity, u, is 0 m/s. The equation simplifies to:
s = (1/2)at^2
Substituting the given values:
1.4 = (1/2)(1.67)t^2
To isolate t^2, we divide both sides of the equation by (1/2)(1.67):
t^2 = 1.4 / [(1/2)(1.67)]
t^2 = 1.4 / 0.835
t^2 ≈ 1.676
To find t, we take the square root of both sides:
t ≈ √1.676
t ≈ 1.296 seconds
Therefore, it takes approximately 1.296 seconds for the feather to fall on the surface of the moon.