Time for Droplet of Water to Fall .25m from Hole 1. The problem statement, all variables and given/known data A cylindrical tank .7m tall is filled with water and placed on a stand (below it) that is .3m tall. A hole of radius .001 m in the bottom of tank is opened. Water then flows through the hole and through an opening in the stand and is collected in a tray .3 m below the hole. At the same time, water is added to the tank at an appropriate rate so that the water level in the tank remains constant. Find: ^The speed at which the water flows out from the hole [Done: 3.7 m/s] ^The volume rate at which water flows out from the hole [Done: 1.1623893 x 10-5 m3/s] ^The volume of water collected in the tray in 2 minutes [Done: .0013948672 m3] ^ ! The time it takes for a droplet of water to fall 0 .25 m from the hole. 2. Relevant equations Density of Water: 1000 kg/m3 3. The attempt - PART D - *The time it takes for a droplet of water to fall 0 .25 m from the hole. ΔX = V°*t + (1/2)*a*t2 - .25 = 3.7t + (1/2)(9.8)t2 - 4.9t2 + 3.7t - 0.25 = 0 - Used Quadratic Equation but answer turned out negative and does not match answers - ? ? ? Correct Answer: 0. 062 seconds My question is how.