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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This note was uploaded on 01/15/2012 for the course PHY 2048 taught by Professor Field during the Fall '08 term at University of Florida.
 Fall '08
 Field
 Physics, Work

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