Question

a feather is dropped on the moon from a height of 2 meters. the acceleration of gravity on the moon is 1.66 m/s/s. determine the time for the feather to fall to the surface of the moon. question 8 10 pts a marble rolling 8.3 m/s rolls off a table that is 2 meters high. how far from the edge of the table does it land?

Explanation:

Step1: Analyze the vertical motion

For the feather on the moon, the initial vertical velocity \(u = 0\ m/s\), the acceleration \(a=1.66\ m/s^{2}\), and the displacement \(s = 2\ m\). Use the equation \(s=ut+\frac{1}{2}at^{2}\). Since \(u = 0\), the equation simplifies to \(s=\frac{1}{2}at^{2}\).

Step2: Solve for time \(t\)

Rearrange \(s=\frac{1}{2}at^{2}\) to get \(t=\sqrt{\frac{2s}{a}}\). Substitute \(s = 2\ m\) and \(a = 1.66\ m/s^{2}\) into the formula: \(t=\sqrt{\frac{2\times2}{1.66}}\).
Calculate \(\frac{2\times2}{1.66}=\frac{4}{1.66}\approx2.4096\). Then \(t=\sqrt{2.4096}\approx1.55\ s\).

For the marble:

Step1: Analyze the vertical motion

The initial vertical velocity \(u_y=0\ m/s\), the acceleration \(a = g=9.8\ m/s^{2}\), and the displacement \(s_y = 2\ m\). Use \(s_y=u_yt+\frac{1}{2}gt^{2}\). Since \(u_y = 0\), \(t=\sqrt{\frac{2s_y}{g}}\). Substitute \(s_y=2\ m\) and \(g = 9.8\ m/s^{2}\) into the formula: \(t=\sqrt{\frac{2\times2}{9.8}}=\sqrt{\frac{4}{9.8}}\approx0.64\ s\).

Step2: Analyze the horizontal motion

The horizontal velocity \(u_x = 8.3\ m/s\) (constant, as there is no horizontal acceleration). Use the formula \(s_x=u_xt\). Substitute \(u_x = 8.3\ m/s\) and \(t\approx0.64\ s\) into the formula: \(s_x=8.3\times0.64 = 5.312\ m\).

Answer:

The time for the feather to fall: \(1.55\ s\).
The distance from the edge of the table for the marble: \(5.31\ m\).