Question

A feather is dropped on the moon from a height of 1.40 meters. The acceleration of gravity on the moon is 1.67 m/s2. Determine the time for the feather to fall to the surface of the moon.

Answer 1

g=2h/t^2
1.67=(2*1.4)/t^2
t^2=280/167
t=1.29 sec

Answer 2

Time Taken = 1.29 Seconds

Given:

Vi = 0 m/sec

Displacement = 1.40 meters

Acceleration = 1.67 m/sec^2

To Find:

Time taken for the feather to fall to the surface of the moon.

Solution:

We use the equation:

d = vi*t + 0.5*a*t^2

Substituting the values in this equation we get:

-1.40 = 0 * t + 0.5 * -1.67 * t^2

-1.40 = 0 + (-0.835) * t^2

-1.40 = (-0.835) * t^2

Bringing -0.835 into the other side:

t^2 = - 1.40 m / -0.835 m / sec^2

t^2 = 1.68 sec^2

Removing the square we get:

Time =  1.29 seconds

Therefore, the time taken for the feather to fall to the surface of the moon is 1.29 seconds.