A swimmer bounces straight up off a diving board with an initial upward velocity of 4.00 m/s, from a takeoff point 8.00 m above the pool, then falls feet first into the water. Using g = 9.80 m/s^2, how long are her feet in the air?

1. Take the pool surface as y = 0, so y_0 = 8.00 m and v_0 = 4.00 m/s upward. 2. Set y = y_0 + v_0 t - 0.5 g t^2 equal to 0: 0 = 8.00 + 4.00t - 4.90t^2. 3. The quadratic formula gives t = (4.00 + square root of 172.8) / 9.80. 4. t = (4.00 + 13.145) / 9.80 = 1.75 s, taking the positive root since time cannot be negative. _Source: OpenStax Physics (CC BY 4.0), Ch 3 "Acceleration", section 3.2 Representing Acceleration with Equations and Graphs_