Question

a coin is dropped into a wishing well. how deep is the well if it hits the bottom with a speed of 27 m/s?

Explanation:

Step1: Identify the kinematic equation

We use the kinematic equation \(v^{2}=v_{0}^{2}+2ah\). Since the coin is dropped, \(v_{0} = 0\) m/s, \(a = g= 9.8\) m/s² (acceleration due to gravity), and \(v = 27\) m/s.

Step2: Solve for \(h\)

Substitute the values into the equation \(v^{2}=v_{0}^{2}+2ah\).

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Answer:

\(h\approx37.2\) m