Question

Water is ejected from the water nozzle with a speed v0 = 49.5 ft/sec. For what value of the angle θ will the water land closest to the wall after clearing the top? Neglect the effects of wall thickness and air resistance. What distance x to the right of point B does the water land?

Chapter 2, Practice Problem 2/036 Water is ejected from the water nozzle with a speed Vo = 49.5 ft/sec. For what value of the angle & will the water land closest to the wall after clearing the top? Neglect the effects of wall thickness and air resistance. What distance x to the right of point B does the water land? 4.9 - 69' Not to scale Answer: X

Answer 1

Imagine water stream as particles, its location in (x,y) planeis, (x,y)=( 69',4.9') Along x axis , speed of water is uniform its acceleration (a = 0). x=v, cos et 69 = (49.5t)cos e - (1) y = y +ve sin 4.9=1.0+(49.51 3.9+'gt? = (49.54 )sin e .... (2) Squaring and adding equations (1) & (2) and simplify: (49.5€)* [ sinº8+ cose] =(59)2 +[ 3.9+berat 2450.254° = 4776.21+2(32.281/82)*+*+3.9(32.26/s*)e2 2324.67 +* = 4776.21+259.214* (here t = x) 259.21x2 - 2324.67 x +4776.21 = 0) -------- (1) 259.21x2 - 2324.67 x +4776.21 = 0 Using Quadratic equation calculator solve x: x = 5.78 (or) 3.187 * = 5.78 (or) 3.187 1 = 2.404 s(or)1.785 s From the equation (1) fort = 2.404 s , 69 = (49.5(2.404 s ))cos 8 + 9 = 54.56 From the equation (1) fort=1.785 s , 69 = (49.5(1.785 s ))cos 8 +9= 38.65) Out of two , 8 = 54.56* *water land closest to the wall. Therefore, 9 = 54.56°54.6" * Answer **

(2) Using 8 = 54.56', time taken by water to reach the land is, 0 =1.0 +(49.5t ) sin 8 - 0 =1.0+(49.5t ) sin 54.56* - }(32.2 ft/s*) (* 16.12 - 40.328t+1=0 Using Quadratic equation calculator solve t: t = 12.4795 Water lands to the right of B is, x = (49.51) cosa = (495/2.479s)) cos(54.56) 71.2'* → Answer**