Unformatted text preview: t/v s ) d + (1 /v 2 s ) d 2 . Now multiply by gv 2 s and rearrange to get gd 2 − 2 v s ( gt + v s ) d + gv 2 s t 2 = 0. This is a quadratic equation for d . Its solutions are d = 2 v s ( gt + v s ) ± p 4 v 2 s ( gt + v s ) 2 − 4 g 2 v 2 s t 2 2 g . The physical solution must yield d = 0 for t = 0, so we take the solution with the negative sign in front of the square root. Once values are substituted the result d = 40 . 7 m is obtained....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics

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