problem03_48

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

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Unformatted text preview: 3.48: a) The equations of motions are: y = h + (v0 sin )t - x = (v0 cos )t v y = v0 sin - gt vx = v0 cos Note that the angle of 36.9o results in sin 36.9 = 3/5 and cos 36.9 = 4/5 . b) At the top of the trajectory, v y = 0 . Solve this for t and use in the equation for y to find the maximum height: t = reduces to y = h + y =h+ ( 25 gh / 8 )( 3 / 5 ) 2 2g v 0 2 sin 2 2g v0 sin g 1 2 gt 2 . Then, y = h + (v 0 sin ) ( v0 sin g ) - g( 1 2 v0 sin 2 g ) , which . Using v0 = 25 gh / 8 , and sin = 3 / 5 , this becomes 25 16 9 = h + 16 h , or y = h . Note: This answer assumes that y0 = h . Taking 9 y0 = 0 will give a result of y = 16 h (above the roof). c) The total time of flight can be found from the y equation by setting y = 0 , assuming y0 = h , solving the quadratic for t and inserting the total flight time in the x equation to 2 3 find the range. The quadratic is 1 gt - 5 v 0 - h = 0 . Using the quadratic formula gives 2 t= ( 3 / 5) v0 ( - ( 3 / 5 ) v0 ) 2 - 4 ( 1 g )( - h ) 2 2( 1 g ) 2 . Substituting v0 = 25 gh / 8 gives t = 1 2 ( 3 / 5) 25 gh / 8 g 9 25 gh 16 gh 8 + 8 25 Collecting terms gives t: t = meaningful and so t = 4 h 2g ( 9h 2g 25 h 2g ) = (3 1 2 . Then, using x = (v0 ) . Only the positive root is ( ) ( 4 ) = 4h . cos )t , x = h 2g . 5 h 2g 25 gh 4 8 5 h 2g ...
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This document was uploaded on 02/04/2008.

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