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?
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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.
