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/s squared. Determine the time for the feather to fall to the surface of the moon? And can you show me how you got the answer.

Question

amistre64 · Answer

$$h=-\frac12gt^2+v_ot+h_o$$

amistre64 · Answer

since its a quadratic equation, use the quadratic formula:$$t=\frac{-v_o\pm\sqrt{(v_o)^2-4(-\frac12g)(h_o)}}{2(-\frac12g)}$$

amistre64 · Answer

which can be simplified as:

$$t=\frac{v_o\pm\sqrt{(v_o)^2+2gh_o}}{g}$$
since the initial velocity vo = 0
$$t=\frac{\sqrt{2gh}}{g}$$