Mark is in a deep hole looking for treasure. He is standing 8 feet below the surface. He throws an old watch he found with an initial upward velocity of 23 ft/sec. How long until it lands outside the hole, having gone up and back down?
He trows a watch with an initial upward velocity: v o = 23 ft/s. v = v o - g · t At the highest point: 0 = 23 - 9.81 · t 9.81 t = 23 t = 23 : 9.81 t = 2.34 s h = v o · t - g t² / 2 h = 23 · 2.34 - 9.81 · 2.34² / 2 h = 53.92 - 26.86 = 26.86 ft 26.86 ft - 8 ft = 18.86 ft 18.86 = g · t² / 2 t² = 18.86 · 2 : 9.81 t = 1.96 s t ( total ) = 2.34 s + 1.96 s = 4.3 s Answer: The watch will land outside the hole after 4.3 seconds.
