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Physics 210 Problems - My Solutions Dr. Hulan E. Jack Jr. Chapter 9 P53 Serway, Faughn and Vuille: College Physics 8th Edition , Thomson Brooks/Cole, Vol I(ISBN #) 978-049511374-3 THE PROBLEM STATEMENT Ch 9 P53. A jet of water squirts out horizontally from a hole near the bottom of the tank shown in Figure P9.53. If the hole has a diameter of 3.50 mm, what is the height h of the water level in the tank? . Page 1 of 2 Physics 210 Problems - My Solutions Dr. Hulan E. Jack Jr. Ch 9 P53. A jet of water squirts out horizontally from a hole near the bottom of the tank shown in Figure P9.53. If the hole has a diameter of 3.50 mm, what is the height h of the water level in the tank? Basic Solution (Including 1BRAINSTORMING-Definitions, concepts , rinciples and Discussion) Given: H= 1.00 m, x = 0.600m . Find: h and vh Note the pressure top , Ptop ,at the hole Phole, and ouside, Poutside, are all the atmospheric pressure, Patm. same value Patm Ptop = Phole = Poutside = Patm . Physical Principles: 1. The stream from hole to ground is in free fall. Horizontal vh = const. , hence x = vht . so t = x / vh . (1) (2) hole Vertical av = -g, so y = v0yt – ½ gt2 , with v0y =0. (3) Hence Eq. (3) gives y = ½ gt2 = H (4) vh H So t= sqrt(2H/g) (5) = sqrt(2*1.00m/9.8m/s2) = 0.452 s x Using Eq. (5) in Eq. (1) gives (6) vh = x/t = x/sqrt(2H/g) = 0.600m/ sqrt(2H/g) = 0.600/0.452s = 1.33 m/s . 2. Bernoulli’s Equation in fluid. Ptop /+ ½ vtop2 + gh = Phole/ + ½ vh2 + 0 and vtop = 0 (7) gives , gh = ½ vh2 = h = vh2 /2g = = = = = = (8) [x/{sqrt(2H/g}]2/(2g) [x2/{(2H/g}]/(2g) = [x2g/{(2H}]/(2g) [x2g/{(2H*2g }] x2/4H 0.6002 m2 /(4*1) = 0.3600/4 = 0.0900m = 9.00 cm Page 2 of 2