Unformatted text preview: y . Thus, using T t ) 5 / 4 ( = and . , ) ( ) ( , 2 25 8 2 25 16 2 5 2 max 2 5 4 2 1 5 4 5 4 2 5 2 max 2 2 1 max gT gT gT y T g T gT gT y gt t v y y y+ =+ =+ = Combining terms, 2 25 18 max gT y = . d) To find the total horizontal distance, break the problem into three parts: The boost phase, the rise to maximum, and the fall back to earth. The fall time back to earth can be found from the answer to part (c), 2 2 ) 2 / 1 ( ) 25 / 18 ( gt gT = , or T t ) 5 / 6 ( = . Then, multiplying these times and the velocity, ) )( ( ) )( ( 5 6 5 3 5 4 5 3 2 10 3 T gT T gT gT x + + = , or 2 25 18 2 25 12 2 10 3 gT gT gT x + + = . Combining terms gives 2 2 3 gT x = ....
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 Vector Space, Euclidean geometry, Line segment

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