document.write( "Question 909476: I need someone who is good in physics i need help please .i do not understand this question.\r
\n" ); document.write( "\n" ); document.write( "An athlete executing a long jump leaves the ground at 28.0 angle and travels 7.80 m a)What was the take off speed? b)If this speed were increased by just 5.0%,how much longer would the jump be?
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Algebra.Com's Answer #551934 by KMST(5328)\"\" \"About 
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The athlete left the ground with an initial velocity of magnitude \"v\" (in m/s), making an angle of \"28%5Eo\" with the ground. The magnitude and direction of the athlete's velocity changed over time under the influence of the acceleration of gravity, \"g=9.6\"\"m%2Fs%5E2\" downwards, until the athlete landed on the ground , \"t\" seconds after leaving the ground.
\n" ); document.write( "The velocity can be decomposed into a horizontal component of magnitude \"v%2Acos%2828%5Eo%29\" ,
\n" ); document.write( "and a vertical component of magnitude \"v%2Asin%2828%5Eo%29\" .
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\n" ); document.write( "FOR THE HORIZONTAL COMPONENT OF THE MOTION:
\n" ); document.write( "We assume that the athlete is jumping over a perfectly horizontal surface, and that air resistance does not really affect him/her at the speeds he/she is capable of attaining.
\n" ); document.write( "That makes his horizontal speed constant at \"v%2Acos%2828%5Eo%29\" for the \"t\" seconds the athlete is airborne.
\n" ); document.write( "As a consequence of our assumptions, we know that
\n" ); document.write( "the athlete's initial horizontal velocity (in m/s)
\n" ); document.write( "times the time in the air (\"t\" , in seconds)
\n" ); document.write( "equals the horizontal displacement, \"d\" in meters (7.8 m, in this case).
\n" ); document.write( "So, in general \"v%2Acos%2828%5Eo%29%2At=d\" ,
\n" ); document.write( "and in this particular case \"v%2Acos%2828%5Eo%29%2At=7.8\" .
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\n" ); document.write( "FOR THE VERTICAL COMPONENT OF THE MOTION:
\n" ); document.write( "The vertical component of the velocity changes from \"v%2Asin%2828%5Eo%29\" to zero over \"t%2F2\" seconds, at a constant rate of \"g=9.8\"\"m%2Fs%5E2\" .
\n" ); document.write( "So, \"v%2Asin%2828%5Eo%29%2F%28t%2F2%29=9.8\"--->\"v%2Asin%2828%5Eo%29=9.8%2A%28t%2F2%29\"--->\"v%2Asin%2828%5Eo%29=4.9t%29\"--->\"t=v%2Asin%2828%5Eo%29%2F4.9%29\"
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\n" ); document.write( "PUTTING IT ALL TOGETHER:
\n" ); document.write( "How are \"v\" and \"d\" related?
\n" ); document.write( "Putting together \"v%2Acos%2828%5Eo%29%2At=d\" and \"t=v%2Asin%2828%5Eo%29%2F4.9%29\" we get
\n" ); document.write( "\"v%2Acos%2828%5Eo%29%2A%28v%2Asin%2828%5Eo%29%2F4.9%29=d\"--->\"d=%28cos%2828%5Eo%29%2Asin%2828%5Eo%29%2F4.9%29%2Av%5E2\"
\n" ); document.write( "Substituting the approximate values \"sin%2828%5Eo%29=0.46947\" and \"cos%2828%5Eo%29=0.88295\" ,
\n" ); document.write( "to get the approximate formula \"d=0.08460v%5E2\"
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\n" ); document.write( "WHAT DOES THAT TELL YOU?:
\n" ); document.write( "a) At this point you can use \"d=0.08460v%5E2\" to calculate \"v\" :
\n" ); document.write( "\"7.8=0.08460v%5E2\"--->\"v%5E2=7.8%2F0.08460\"--->\"v%5E2=92.20\"--->\"v=sqrt%2892.20%29\"-->\"v=9.6\" (rounded).
\n" ); document.write( "So, the initial velocity was \"highlight%289.6%29\" m/s.
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\n" ); document.write( "b) If an athlete jumps at the same angle, with \"2\" times the initial velocity, he/she will get \"2%5E2=4\" times as far.
\n" ); document.write( "If the athlete jumps with \"1.05\" times (5% more) initial velocity,
\n" ); document.write( "he/she will get \"1.05%5E2=1.1025\" times farther.
\n" ); document.write( "That would be \"1.1025%2A7.8=8.6\" (rounded).
\n" ); document.write( "So, jumping with 5% higher initial velocity, the athlete will travel 8.7 meters.
\n" ); document.write( "That means that his/her jump would be
\n" ); document.write( "8.6 m - 7.8 m = \"highlight%280.8%29\" m longer,
\n" ); document.write( "or 1.10 times longer (10% longer).
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