qz9sol_6839sp10 - 2 t= [2(H-h)/g] The horizontal...

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TA: Tomoyuki Nakayama Tuesday, April 6 th , 2010 PHY 2048: Physic 1, Discussion Section 6839 Quiz 9 (Homework Set # 12) Name: UFID: Formula sheets are not allowed. Calculators are allowed. Do not store equations in your calculator. You need to show all of your work for full credit. ________________________________________________________________________________ The figure below right shows a stream of water flowing through a hole at depth h = 25.0 cm in a tank holding water to height H = 80.0 cm. a) At what distance x does the stream strike the floor? Applying Bernoulli’s equation at the top surface of water and at the hole, we get p 0 + ρgh = p 0 + (1/2)ρv 2 v = √(2gh) Here we ignored the velocity of the water at the top surface because it is very small compared to the velocity at the hole. The motion of the ejected water is considered a projectile motion. The water stream falls down a vertical distance of H – h before it hits the floor. The flight time is H-h = (1/2)gt
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Unformatted text preview: 2 t= [2(H-h)/g] The horizontal displacement is x = vt = (2gh) [2(H-h)/g] = 2(-h 2 +Hh) = 74.2 cm b) At what depth should a second hole be made to give the same value of x ? Let the depth of the new hole be h. This hole gives the same value of x as h, thus we obtain 2(-h 2 +Hh) = 2(-h 2 +Hh) -h 2 +Hh = -h 2 +Hh h 2 - h 2 + Hh Hh = 0 (h+h - H)(h h ) = 0 h = h, H- h The first one is trivial, the second one yields H- h = 80 25 =- 55.0 cm. c) At what depth should a hole be made to maximize x ? We already have the expression for x (=2(-h 2 +Hh) ). If the function inside of the square root is maximum, then x is maximum too. The function takes its maximum value at a point where its first derivative (the slope of the graph) is zero. Setting the derivative of the function to be zero, we get f(h) =-h 2 +Hh f(h) = -2h + H = 0 h = H/2 = 40.0 cm...
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