33. A rock having a mass of 1.25 kg is dropped from the edge of a clill to the surface of the river water 35.0 m below. With what speed does the rock strike the surface of the water? A. 20.7m/s B. 26.2m/s C. . 172m/s D. . 442m/s

Here's how to solve this problem using the principles of conservation of energy:<br /><br />**1. Understand the concept:**<br /><br />The rock initially has potential energy due to its height above the water. As it falls, this potential energy is converted into kinetic energy (energy of motion). We can assume negligible air resistance.<br /><br />**2. Set up the equation:**<br /><br />Potential Energy (PE) = Kinetic Energy (KE)<br /><br />* PE = mgh (where m is mass, g is acceleration due to gravity (9.8 m/s²), and h is height)<br />* KE = (1/2)mv² (where m is mass and v is velocity)<br /><br />Therefore: mgh = (1/2)mv²<br /><br />**3. Solve for velocity (v):**<br /><br />Notice that the mass (m) cancels out on both sides of the equation:<br /><br />gh = (1/2)v²<br /><br />2gh = v²<br /><br />v = sqrt(2gh)<br /><br />**4. Plug in the values:**<br /><br />v = sqrt(2 * 9.8 m/s² * 35.0 m)<br />v = sqrt(686 m²/s²)<br />v ≈ 26.2 m/s<br /><br />**Answer:**<br /><br />B. 26.2 m/s<br />