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Unformatted text preview: 2 t= [2(Hh)/g] The horizontal displacement is x = vt = (2gh) [2(Hh)/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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Document
 Fall '08
 Field
 Physics, Work

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