problem02_98 solution

University Physics with Modern Physics with Mastering Physics (11th Edition)

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2.98: a) Let the height be h and denote the 1.30-s interval as t ; the simultaneous equations h = 2 2 1 3 2 2 2 1 ) ( , t t g h gt - = can be solved for t. Eliminating h and taking the square root, , and , 3 2 1 2 3 - - = = t t t t t and substitution into 2 2 1 gt h = gives h = 246 m. This method avoids use of the quadratic formula; the quadratic formula is a generalization of the method of “completing the square”, and in the above form, , ) ( 2 2 1 3 2 t t g h - = the square is already completed. b) The above method assumed that
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Unformatted text preview: t > 0 when the square root was taken. The negative root (with ∆ t = 0) gives an answer of 2.51 m, clearly not a “cliff”. This would correspond to an object that was initially near the bottom of this “cliff” being thrown upward and taking 1.30 s to rise to the top and fall to the bottom. Although physically possible, the conditions of the problem preclude this answer. Page 1 of 1...
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This document was uploaded on 02/04/2008.

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