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?

To solve this problem, we can use the kinematic equation that relates initial velocity, final velocity, acceleration due to gravity, and displacement (height in this case). The equation is: \[ v^2 = u^2 + 2as \] Where: - \( v \) = final velocity (86 m/s) - \( u \) = initial velocity (0 m/s since the coyote is dropped) - \( a \) = acceleration (9.81 m/s²) - \( s \) = height of the cliff (what we want to find) Since the coyote is dropped, the initial velocity \( u = 0 \), so the equation simplifies to: \[ v^2 = 2as \] Rearranging gives us: \[ s = \frac{v^2}{2a} \] Now, substituting the values: \[ s = \frac{(86 \, \mathrm{m/s})^2}{2 \times 9.81 \, \mathrm{m/s}^2} \] Calculating the values: \[ s = \frac{7396}{19.62} \approx 377.38 \, \mathrm{m} \] Therefore, the height of the cliff is approximately 377.38 meters.