A girl is standing at the edge of a cliff \(100 . \mathrm{m}\) above the ground. She reaches out over the edge of the cliff and throws a rock straight upward with a speed \(8.00 \mathrm{~m} / \mathrm{s}\). a) How long does it take the rock to hit the ground? b) What is the speed of the rock the instant before it hits the ground?

First, let's consider the initial conditions and the equations: the initial height \(y_0 = 100\) meters, the initial velocity \(v_0 = 8.00 \thinspace m/s\) (upward, so positive), and the acceleration due to gravity \(g = 9.81 \thinspace m/s^2\) (downward, so negative). Note that the position equation is \(y(t) = y_0 + v_0t - \frac{1}{2}gt^2\). Since we want to find the time it takes for the rock to hit the ground, we need to find the time \(t\) when \(y(t) = 0\): \(0 = 100 + 8t - \frac{1}{2}(9.81)t^2\) Now, we have a quadratic equation of the form \(0 = at^2 + bt + c\), where \(a = -\frac{1}{2}(9.81)\), \(b = 8\), and \(c = 100\). Solve for \(t\) using the quadratic formula, \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Plugging in our values, we have: \(t = \frac{-8 \pm \sqrt{8^2 - 4(-\frac{1}{2}(9.81))(100)}}{2(-\frac{1}{2}(9.81))}\) Solving for \(t\), we get two possible values, \(t \approx 1.83\) seconds or \(t \approx 11.1\) seconds. Since the first value represents the time when the rock is at ground level on its way up, we only consider the second value: \(t \approx 11.1\) seconds for the rock to hit the ground.

Now that we have the time when the rock hits the ground, we can find its vertical velocity before it hits the ground. We'll use the equation \(v(t) = v_0 - gt\). Substitute the values for \(v_0\), \(g\), and the time \(t \approx 11.1\) seconds: \(v(t) = 8 - (9.81)(11.1)\) \(v(t) \approx -109 \thinspace m/s\) The vertical velocity before the rock hits the ground is approximately \(109 \thinspace m/s\) downward. Note that the final speed of the rock does not depend on its initial height and only depends on the initial speed and the time it takes to hit the ground.