427_Physics ProblemsTechnical Physics

427_Physics ProblemsTechnical Physics - Chapter 14...

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Chapter 14 429 Additional Problems P14.54 Consider the diagram and apply Bernoulli’s equation to points A and B, taking y = 0 at the level of point B, and recognizing that v A is approximately zero. This gives: P g h L P v g A w w B w B w + + = + + 1 2 0 1 2 0 2 2 ρ ρ θ ρ ρ a f a f a f sin Now, recognize that P P P A B = = atmosphere since both points are open to the atmosphere (neglecting variation of atmospheric pressure with altitude). Thus, we obtain h A Valve L B θ FIG. P14.54 v g h L v B B = = ° = 2 2 9 80 10 0 2 00 30 0 13 3 sin . . . sin . . θ a f e j a f m s m m m s 2 Now the problem reduces to one of projectile motion with v v yi B = °= sin . . 30 0 6 64 m s. Then, v v a y yf yi 2 2 2 = + b g gives at the top of the arc (where y y = max and v yf = 0) 0 6 64 2 9 80 0 2 = + . . max m s m s 2 b g e j b g y or y max . = 2 25 m above the level where the water emerges b g . P14.55
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