document.write( "Question 765572: Please help solve this. It would help me a lot in reviewing for my final exam. Thank you so much!\r
\n" ); document.write( "\n" ); document.write( "A rubber ball is dropped from a height of 12 feet. If it rebounds one-third of the distance it has fallen after each fall, a) how far will it rebound the 8th time? b) through what distance has it traveled when it strikes the ground the 10th time?
\n" ); document.write( " \n" ); document.write( "

Algebra.Com's Answer #466312 by ramkikk66(644) $\"\"$ $\"About$

You can put this solution on YOUR website!


\n" );
document.write( "The ball bounces to a 1/3 its initial height every time, which means the series of heights forms a geometric progression, with common ratio as 1/3
\n" );
document.write( "First term a = Initial height = 12
\n" );
document.write( "2nd term = 1st rebound = 12*1/3 = 4
\n" );
document.write( "3rd term = 2nd rebound = 4*1/3 = 4/3\r
\n" );
document.write( "\n" );
document.write( "So the height of the 8th rebound is given by the 9th term in the series.
\n" );
document.write( "9th term = a*r^(n-1) = 12*(1/3)^8 = 12/6561 = 4/2187\r
\n" );
document.write( "\n" );
document.write( "Height of rebound for 8th time = 4/2187.\r
\n" );
document.write( "
\n" );
document.write( "\n" );
document.write( "2) Total distance travelled at the 10th time it hits ground is the sum of  the 
\n" );
document.write( "upward travel and the downward travels. Both are geometric progressions.\r
\n" );
document.write( "\n" );
document.write( "Downward travel:
\n" );
document.write( "1st term = 12
\n" );
document.write( "Common ratio = 1/3
\n" );
document.write( "Sum of 10 terms = 
\n" );
document.write( "\r
\n" );
document.write( "\n" );
document.write( "\r
\n" );
document.write( "\n" );
document.write( "= \r
\n" );
document.write( "\n" );
document.write( "= \r
\n" );
document.write( "\n" );
document.write( "2) Upward (rebound) travel\r
\n" );
document.write( "\n" );
document.write( "1st term = 4
\n" );
document.write( "Common ratio = 1/3
\n" );
document.write( "Sum of 9 terms = (since it rebounds 9 times)
\n" );
document.write( "\r
\n" );
document.write( "\n" );
document.write( "\r
\n" );
document.write( "\n" );
document.write( "= \r
\n" );
document.write( "\n" );
document.write( "= \r
\n" );
document.write( "\n" );
document.write( "So the total travel = 17.999 + 5.999 = 23.998 approximately\r
\n" );
document.write( "\n" );
document.write( "As you would see, the sum is already approaching the \"sum to infinity\" of the
\n" );
document.write( "series.\r
\n" );
document.write( "
\n" );
document.write( "\n" );
document.write( ":)\r
\n" );
document.write( "
\n" );
document.write( "
\n" );
document.write( "\n" );
document.write( " \n" );
document.write( "

\n" );