Question

1. a feather, held by an astronaut, is dropped on the moon from a height of 1.40 m. the acceleration of gravity on the moon is 1.67 m/s². determine the time it took for the feather to fall to the surface of the moon.

Explanation:

Step1: Identify the kinematic equation

We use the kinematic equation for free - fall (since the feather is dropped, initial velocity \(u = 0\ m/s\)): \(h=ut+\frac{1}{2}at^{2}\), where \(h\) is the height, \(u\) is the initial velocity, \(a\) is the acceleration, and \(t\) is the time. Given \(u = 0\ m/s\), the equation simplifies to \(h=\frac{1}{2}at^{2}\).

Step2: Rearrange the equation for \(t\)

From \(h=\frac{1}{2}at^{2}\), we can solve for \(t\). First, multiply both sides by \(2\) to get \(2h = at^{2}\). Then divide both sides by \(a\): \(t^{2}=\frac{2h}{a}\). Take the square root of both sides: \(t=\sqrt{\frac{2h}{a}}\).

Step3: Substitute the values

We know that \(h = 1.40\ m\) and \(a=1.67\ m/s^{2}\). Substitute these values into the formula for \(t\):

\(t=\sqrt{\frac{2\times1.40}{1.67}}\)

First, calculate the numerator: \(2\times1.40 = 2.80\)

Then, calculate the fraction: \(\frac{2.80}{1.67}\approx1.6766\)

Then take the square root: \(t=\sqrt{1.6766}\approx1.295\ s\approx1.30\ s\)

Answer:

The time it takes for the feather to fall is approximately \(\boldsymbol{1.30\ s}\) (or more precisely, approximately \(1.29\ s\) to \(1.30\ s\) depending on the level of precision in calculation).