problem03_73

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

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3.73 : a) The acceleration is given as g at an angle of ° 1 . 53 to the horizontal. This is a 3- 4-5 triangle, and thus, g a g a y x ) 5 / 4 ( and ) 5 / 3 ( = = during the "boost" phase of the flight. Hence this portion of the flight is a straight line at an angle of ° 1 . 53 to the horizontal. After time T , the rocket is in free flight, the acceleration is 0 = x a and g a y = , and the familiar equations of projectile motion apply. During this coasting phase of the flight, the trajectory is the familiar parabola. b) During the boost phase, the velocities are: gt v x ) 5 / 3 ( = and gt v y ) 5 / 4 ( = , both straight lines. After T t = , the velocities are gT v x ) 5 / 3 ( = , a horizontal line, and ) ( ) 5 / 4 ( T t g gT v y - - = , a negatively sloping line which crosses the axis at the time of the maximum height. c) To find the maximum height of the rocket, set 0 = y v , and solve for t , where 0 = t when the engines are cut off, use this time in the familiar equation for
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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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This document was uploaded on 02/04/2008.

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