document.write( "Question 1011969: The equation of the path of a cricket ball thrown at an angle of 45 degrees to the horizontal is y = x - x^2/50, where x and y are the horizontal distance travelled and the vertical height respectively. Calculate the greatest vertical height reached and the horizontal distance travelled.\r
\n" ); document.write( "\n" ); document.write( "What i have done so far:\r
\n" ); document.write( "\n" ); document.write( "y = x - x^2/50
\n" ); document.write( "50y = 50x - x^2
\n" ); document.write( " = -(x^2 - 50x) + 0
\n" ); document.write( " = - (x-25)^2 + 625\r
\n" ); document.write( "\n" ); document.write( "Therefore greatest height would be 625 divided by 50 which is 12.5, and that answer is correct.\r
\n" ); document.write( "\n" ); document.write( "But I don't know how to get horizontal distance???
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Algebra.Com's Answer #627815 by macston(5194)\"\" \"About 
You can put this solution on YOUR website!
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\n" ); document.write( "When the height (y)=0, the ball is at minimum or maximum distance
\n" ); document.write( ".
\n" ); document.write( "\"y=x-x%5E2%2F50\"
\n" ); document.write( "Let y=0:
\n" ); document.write( "\"0=x-x%5E2%2F50\"
\n" ); document.write( "\"0=50x-x%5E2\"
\n" ); document.write( "\"0=%28x%29%2850-x%29\"
\n" ); document.write( "\"x=0\" \"OR\" \"50-x=0\"
\n" ); document.write( "\"x=0\" \"OR\" \"50=x\"
\n" ); document.write( "The minimum horizontal distance (start) is 0,
\n" ); document.write( "the maximum horizontal distance (ball hits ground) is 50.
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