Select the correct answer. A bottle lying on the windowsill falls off and takes 4.95 seconds to reach the ground. The distance from the windowsill to the ground is 120.00 meters. Find the time the bottle would take to land if it were to fall the same distance on the moon instead of Earth. (Note Acceleration due to gravity on the moon is 1/6 that on Earth.) A. 0.825 seconds B. 4.08 seconds C. 12.1 seconds D. 29.7 seconds E.146.94 seconds

To find the time the bottle would take to land if it were to fall the same distance on the moon instead of Earth, we can use the formula for the time of free fall:<br /><br />\[ t = \sqrt{\frac{2h}{g}} \]<br /><br />where:<br />- \( t \) is the time of free fall,<br />- \( h \) is the height (distance from the windowsill to the ground),<br />- \( g \) is the acceleration due to gravity.<br /><br />Given:<br />- \( h = 120.00 \) meters,<br />- \( g_{\text{moon}} = \frac{1}{6} g_{\text{earth}} \).<br /><br />First, we need to find the acceleration due to gravity on Earth (\( g_{\text{earth}} \)). We can use the time and distance provided for the fall on Earth:<br /><br />\[ t_{\text{earth}} = 4.95 \text{ seconds} \]<br />\[ h = 120.00 \text{ meters} \]<br /><br />Using the formula for the time of free fall on Earth:<br /><br />\[ t_{\text{earth}} = \sqrt{\frac{2h}{g_{\text{earth}}}} \]<br /><br />Solving for \( g_{\text{earth}} \):<br /><br />\[ 4.95 = \sqrt{\frac{2 \times 120.00}{g_{\text{earth}}}} \]<br />\[ 4.95^2 = \frac{240.00}{g_{\text{earth}}} \]<br />\[ g_{\text{earth}} = \frac{240.00}{4.95^2} \]<br />\[ g_{\text{earth}} = \frac{240.00}{24.75} \]<br />\[ g_{\text{earth}} \approx 9.70 \text{ m/s}^2 \]<br /><br />Now, we can find the acceleration due to gravity on the moon:<br /><br />\[ g_{\text{moon}} = \frac{1}{6} g_{\text{earth}} \]<br />\[ g_{\text{moon}} = \frac{1}{6} \times 9.70 \]<br />\[ g_{\text{moon}} \approx 1.63 \text{ m/s}^2 \]<br /><br />Next, we use the formula for the time of free fall on the moon:<br /><br />\[ t_{\text{moon}} = \sqrt{\frac{2h}{g_{\text{moon}}}} \]<br />\[ t_{\text{moon}} = \sqrt{\frac{2 \times 120.00}{1.63}} \]<br />\[ t_{\text{moon}} = \sqrt{\frac{240.00}{1.63}} \]<br />\[ t_{\text{moon}} = \sqrt{147.24} \]<br />\[ t_{\text{moon}} \approx 12.1 \text{ seconds} \]<br /><br />Therefore, the correct answer is:<br /><br />C. 12.1 seconds