\( \left.\begin{array}{l}\text { 28. An object is thrown off of a cliff } 320 \mathrm{~m} \text { above level } \\ \text { ground with an initial horizontal velocity of } 20 \mathrm{~m} / \mathrm{s} \text {. } \\ \text { The amount of time it takes to strike the ground is } \\ \text { most nearly } \\ \begin{array}{llll}\text { A) } 2 \mathrm{~s} & \text { B) } 4 \mathrm{~s} & \text { C) } 8 \mathrm{~s} & \text { D) } 12 \mathrm{~s}\end{array} \\ \text { E) } 16 \mathrm{~s}\end{array} \begin{array}{llll}\text { 29. } & \text { D } 5.2 \mathrm{~s} & \text { D) } 6.0 \mathrm{~s} & \text { E) } 7.8 \mathrm{~s}\end{array}\right] \)

To solve how long it takes for the object to hit the ground, we can use the equation of motion for free fall. The formula is \( d = \frac{1}{2} g t^2 \), where \( d \) is the distance fallen (320 m), \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)), and \( t \) is the time in seconds. Rearranging the equation, we find \( t = \sqrt{\frac{2d}{g}} \). Plugging in the values gives us \( t \approx \sqrt{\frac{2 \times 320}{9.81}} \approx 8.1 \, \text{s} \), which corresponds closely to option C) 8 s. For a clearer visual, think of it like timing a roller coaster drop! Just as gravity pulls the coaster down with increasing speed, this object free-falls from the cliff, taking about 8 seconds to reach the ground below while maintaining its horizontal speed of 20 m/s. So not only does it fall, but it also travels horizontally, making a neat trajectory!