ch04-p051

# ch04-p051 - 51. We adopt the positive direction choices...

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we see the first equation gives t = x / v 0 cos θ 0 , and when this is substituted into the second the result is yx gx v =− tan cos . 0 2 0 22 0 2 One may solve this by trial and error: systematically trying values of 0 until you find the two that satisfy the equation. A little manipulation, however, will give an algebraic solution: Using the trigonometric identity 1 / cos 2 0 = 1 + tan 2 0 , we obtain 1 2 1 2 0 2 0 2 2 00 2 0 2 gx v xy gx v tan tan θθ −+ + = which is a second-order equation for tan 0 . To simplify writing the solution, we denote () ( ) 2 11 0 / 9.80 m/s 50 m / 25 m/s 19.6m. cg x v == = Then the second-order equation becomes c tan 2 0 x tan 0 + y + c = 0. Using the quadratic formula, we obtain its solution(s). ( ) 0 4 50 m (50 m) 4 3.44 m 19.6 m 19.6 m tan . 2 2 19.6 m xx y c c c ±−+ ± + The two solutions are given by tan 0 = 1.95 and tan 0 = 0.605. The corresponding (first- quadrant) angles are 0 = 63° and 0 = 31°. Thus,
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## This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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