Question

A thin stream of water flows smoothly from a faucet and falls straight down. at one point the water is flowing at a speed of v1 = 1.71 m/s. at a lower point, the diameter of the stream has decreased by a factor of 0.805. what is the vertical distance h between these two points?

Answer 1

The formulas are, v1d1² = v2d2² ........ (1) h = (v2²-v1²)/2g ...... (2) Given that, v1 = 1.71 m/s we assume that the stream has decreased by a factor d2 =0.805d1 then, v1d1² = v2 (0.805d1)² cancelled both side d1² then we get, v1 = v2 (0.805)² v1 = v2 (0.648025) Sub v1 = 1.71, 1.71 = v2 (0.648025) v2 = 1.71/0.648025 v2 = 2.638787083831642 v2 = 2.64 m/s The vertical distance formula, h = (v2²-v1²)/2g We know that value of gravity constant is 9.8 m/s² h = {(2.64)² - (1.71)²)/2(9.8) h = {(6.9696) - (2.9241)}/19.6 h = (4.0455)/19.6 h = 0.2064030612244898 h = 0.21 cm Therefore, the vertical distance h = 0.21 cm.