Unformatted text preview: 2 3.55: a) With 0 = 45 , Eq. (3.27) is solved for v0 = gx 2 x y . In this case, y = 0.9 m is the change in height. Substitution of numerical values gives v0 = 42.8 m/s . b) Using the above algebraic expression for v0 in Eq. (3.27) gives x y = x 188 m (188.9 m) Using x = 116 m gives y = 44.1 m above the initial height, or 45.0 m above the ground, which is 42.0 m above the fence.
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 Algebra, numerical values, algebraic expression, initial height

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