document.write( "Question 1201879: A ball is given an initial velocity of 10.0 m/s at 37° above the horizontal. Assume that the ball landed at the same height.
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Algebra.Com's Answer #836430 by ikleyn(52830)\"\" \"About 
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\n" ); document.write( "A ball is given an initial velocity of 10.0 m/s at 37° above the horizontal.
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document.write( "(a)  the horizontal component is  \"v%5Bhor%5D\" = 10*cos(37°) = 10*0.7986 = 7.986 m/s.\r\n" );
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document.write( "     Horizontal component of velocity remains the same, with no change,\r\n" );
document.write( "     during the entire flight until landing.\r\n" );
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document.write( "     Vertical component of the initial velocity is  \"v%5Bvert%5D\" = 10*sin(37°) = 10*0.6018 = 6.018 m/s.\r\n" );
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document.write( "(b)  With the found vertical initial velocity in (a), the algebraic function for the height is\r\n" );
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document.write( "          h(t) = \"-4.9t%5E2+%2B+v%5Bvert%5D%2At+%2B+h%5B0%5D\" = \"-4.9t%5E2+%2B+6.018%2At+%2Bh%5B0%5D\".    (1)\r\n" );
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document.write( "     Equation to find the time of landing is\r\n" );
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document.write( "          \"-4.9t%5E2+%2B+6.018%2At+%2Bh%5B0%5D\" = \"h%5B0%5D\".\r\n" );
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document.write( "     After canceling  \"h%5B0%5D\"  in both sides, this equation takes the form\r\n" );
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document.write( "          \"-4.9t%5E2+%2B+6.018%2At\" = 0,\r\n" );
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document.write( "          (-4.9t + 6.018)*t = 0\r\n" );
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document.write( "     and gives the value for the time of landing  t = \"6.018%2F4.9\" = 1.228 of a second.\r\n" );
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document.write( "     So, the ball flights during 1.228 seconds.\r\n" );
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document.write( "(c)  Half of this time,  \"1.228%2F2\" = 0.614 of a second, the ball is on ascending trajectory.\r\n" );
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document.write( "     Second half, or 0.614 of a second, the ball is on descending trajectory.\r\n" );
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document.write( "     To find the range (the horizontal distance), multiply the horizontal component of the velocity \r\n" );
document.write( "     by the flight time\r\n" );
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document.write( "         the range = 7.986*1.228 = 9.807 m (rounded).\r\n" );
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document.write( "     To find the maximum height, find the maximum of this quadratic function (1)  (see above).\r\n" );
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document.write( "     As I just said, during 0.614 of a second the ball moves up, so substitute t= 0.614 s\r\n" );
document.write( "     into the function.\r\n" );
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document.write( "     Doing this way, you will find the maximum height over \"h%5B0%5D\"\r\n" );
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document.write( "         \"h%5Bmax%5D\" = \"-4.9%2A0.614%5E2+%2B+6.018%2A0.614\" = 1.85 m  (rounded).\r\n" );
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