document.write( "Question 1178898: Grade 10- The height of a ball thrown upward after a given amount of time is h = - 4.9t ^ 2 + 29.4t + 1 where h represents the height of the ball in metres , and t represents the time elapsed since the ball has been thrown . Determine the maximum height of the ball and at what time this occurs. What is the maximum height reached by the ball? How long is the ball above a height of 40 m ? When does the ball hit the ground? \n" ); document.write( "
Algebra.Com's Answer #808352 by Boreal(15235)![]() ![]() You can put this solution on YOUR website! the vertex t value is at -b/2a or -29.4/-9.8 or 3 seconds \n" ); document.write( "the maximum height is at h(3)=-44.1+88.2+1=45.1 m \n" ); document.write( "solve for h=40 \n" ); document.write( "so -4.9t^2+29.4t+1=40 \n" ); document.write( "and -4.9t^2+29.4t-39=0 \n" ); document.write( "or 4.9t^2-29.4t+39=0 \n" ); document.write( "t=(1/9.8)(29.4+/-sqrt (99.96); sqrt term=10.0 \n" ); document.write( "t=1.98 sec and t=4.02 sec \n" ); document.write( "it is above 40 m for 2.04 sec \n" ); document.write( "it hits 0 at -4.9t^2+29..4t-1=0=4.9t^2-29.4t+1 \n" ); document.write( "the positive root is t=(1/9.8)*(29.4+sqrt883.86); sqrt term=29.73 \n" ); document.write( "t=6.03 sec\r \n" ); document.write( "\n" ); document.write( " |