Engr240-HW8

# X0 y0 vy 0 vy v y 1 vy vertical motion 0 1 v0 1

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Unformatted text preview: ximum height where t2 (v y )1 g v0 [(1 e) cos 2 2g (1 e)] (5) PROBLEM 13.173 (Continued) Horizontal motion: vx (vx )1 x (vx )1 t x2 1 v0 (1 e)sin 2 2 (vx )1 t2 At the point of maximum height, Let 2 . To determine the value of 2 v0 (1 e)sin 2 [(1 e) cos 2 4g (1 e)] that maximizes x2 , differentiate the expression v with respect to . v sin [(1 e) cos (1 e)] dv d cos [(1 e) cos (1 e)] (1 e)sin 2 ] (1 e) cos 2 (1 e)(1 cos 2 ) (1 e) cos 2(1 e) cos 2 (1 e) cos (1 e) 0 0 This is a quadratic equation for cos . (a) e1 4cos 2 20 cos 2 1 2 2 2 cos 45 and 135 22.5 and 67.5 Reject the negative values which make x2 negative. Reject 67.5 since it makes a sma...
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## This note was uploaded on 12/01/2012 for the course ENGR 240 at Canada College.

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