Old Faithful geyser in Yellowstone Park shoots water hourly to a height of 40 m . With what velocity does the water leave the ground? 1) \( 7.0 \mathrm{~m} / \mathrm{s} \). 2) \( 14 \mathrm{~m} / \mathrm{s} \) 3) \( 20 \mathrm{~m} / \mathrm{s} \) 4) \( 28 \mathrm{~m} / \mathrm{s} \) 5) \( 34 \mathrm{~m} / \mathrm{s} \)

To calculate the velocity at which the water leaves the ground, we can use the kinematic equation for an object in free fall: \[ v^2 = u^2 + 2as \] Here, \(v\) is the final velocity (which is 0 at the peak height), \(u\) is the initial velocity (what we’re looking for), \(a\) is the acceleration due to gravity (approximately \(-9.81 \, \text{m/s}^2\)), and \(s\) is the height (\(40 \, \text{m}\)). Rearranging the equation for initial velocity \(u\): \[ u = \sqrt{v^2 - 2as} \] Substituting \(v = 0\), \(a = -9.81 \, \text{m/s}^2\), and \(s = 40 \, m\): \[ u = \sqrt{0 - 2 \times (-9.81) \times 40} \] \[ u = \sqrt{2 \times 9.81 \times 40} \] Calculating this yields: \[ u \approx \sqrt{784.8} \approx 28 \, \text{m/s} \] Thus, the correct answer is option 4) \(28 \mathrm{~m} / \mathrm{s}\).