Chapter 3, Problem 71A

To determine

The time taken by the feather to hit the Moon’s surface.

Answer to Problem 71A

  $1.22\text{s}$

Explanation of Solution

Given:

A feather is dropped from a height of 1.2m above the surface of the Moon where the free-fall acceleration is $1.62{\text{m/s}}^{\text{2}}$ .

Formula used:

The distance traveled by an object with a constant acceleration is given by,

  $d_f=d_i+v_i\Delta t+\frac{1}{2}a{\left(\Delta t\right)}^2$

Here, d represents displacement, v represents velocity, t represents time and a is the acceleration.

Calculation:

Since, the feather is dropped, thus its initial velocity will be 0, that is,

  $v_i=0$

The free fall acceleration on the Moon is $1.62{\text{m/s}}^{\text{2}}$ . Thus,

  $a=1.62{\text{m/s}}^{\text{2}}$

Now, using the above-mentioned formula for the distance covered, the time elapsed of the motion will be:

  $\begin{array}{l}{d}_f=d_i+v_i\Delta t+\frac{1}{2}a{\left(\Delta t\right)}^2\\ 1.2=0+0\left(\Delta t\right)+\frac{1}{2}\left(1.62\right){\left(\Delta t\right)}^2\\ \Delta t=1.22\text{s}\end{array}$

Conclusion:

Thus, the time taken by the feather to hit the Moon’s surface is $1.22\text{s}$ .