document.write( "Question 30664: A boy throws a rock straight up into the air. It reaches the highest point of its flight after 2.5 seconds. How fast was the rock giong when it left the boys hand? \n" ); document.write( "
Algebra.Com's Answer #17643 by venugopalramana(3286)![]() ![]() You can put this solution on YOUR website! THE FORMULA TO BE USED IS \n" ); document.write( "V=U+AT...WHERE \n" ); document.write( "V=FINAL VELOCITY....FT/SEC. \n" ); document.write( "U=INITIAL VELOCITY....FT/SEC. \n" ); document.write( "A = ACCELERATION ..HERE IT IS DUE TO GRAVITY WHICH IS ACTING AGAINST THE MOTION ..ITS VALUE IS -32 FT/SEC.SEC. \n" ); document.write( "T=TIME OF TRAVEL...SEC \n" ); document.write( "WE HAVE U=? \n" ); document.write( "V=0....AS IT REACHED THE HIGHEST POINT AND WILL STOP AND FALL BACK FROM THERE. \n" ); document.write( "T=2.5 SEC. \n" ); document.write( "0=U-32*2.5 \n" ); document.write( "U=32*2.5=80 FT/ SEC. \n" ); document.write( "IF YOU ARE IN MKS UNITS THEN A=-9.8 M/SEC.SEC \n" ); document.write( "U=9.8*2.5=24.5 M/SEC \n" ); document.write( " |