47. A coyote with a mass of 165 kg is dropped off a cliff. It hits the ground with a speed of \( 86 \mathrm{~m} / \mathrm{s} \). How high was the cliff that it fell from? \( m= \).

To determine the height of the cliff, we can use the principle of conservation of energy. The potential energy at the height \( h \) converts into kinetic energy just before the coyote hits the ground. The equation is: \[ PE = KE \implies mgh = \frac{1}{2} mv^2 \] The mass (\( m \)) cancels out: \[ gh = \frac{1}{2} v^2 \] Solving for height \( h \): \[ h = \frac{v^2}{2g} \] Substituting \( v = 86 \, \text{m/s} \) and \( g \approx 9.81 \, \text{m/s}^2 \): \[ h = \frac{(86)^2}{2 \times 9.81} \approx \frac{7396}{19.62} \approx 377.5 \, \text{m} \] So, the coyote fell from a height of approximately **377.5 meters**!