P15_054 - 54(a Since Sample Problem 15-9 deals with a...

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Unformatted text preview: 54. (a) Since Sample Problem 15-9 deals with a similar situation, we use the final equation (labeled “Answer”) from it: v = 2gh =⇒ v = vo for the projectile motion. The stream of water emerges horizontally (θo = 0◦ in the notation of Chapter 4), and setting y − yo = −(H − h) in Eq. 4-22, we obtain the “time-of-flight” t= −2(H − h) = −g 2 (H − h) . g Using this in Eq. 4-21, where xo = 0 by choice of coordinate origin, we find x = vo t = 2gh 2 (H − h) = 2 h(H − h) . g (b) The result of part (a) (which, when squared, reads x2 = 4h(H − h)) is a quadratic equation for h once x and H are specified. Two solutions for h are therefore mathematically possible, but are they both physically possible? For instance, are both solutions positive and less than H ? We employ the quadratic formula: √ x2 H ± H 2 − x2 2 h − Hh + = 0 =⇒ h = 4 2 which permits us to see that both roots are physically possible, so long as x < H . Labeling the larger root h1 (where the plus sign is chosen) and the smaller root as h2 (where the minus sign is chosen), then we note that their sum is simply √ √ H + H 2 − x2 H − H 2 − x2 h1 + h2 = + =H . 2 2 Thus, one root is related to the other (generically labeled h and h) by h = H − h. (c) We wish to maximize the function f = x2 = 4h(H − h). We differentiate with respect to h and set equal to zero to obtain df H = 4H − 8h = 0 =⇒ h = dh 2 as the depth from which an emerging stream of water will travel the maximum horizontal distance. ...
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