document.write( "Question 1036786: The height of a projectile t seconds after being launched with an initial
\n" ); document.write( "velocity of 96 ft/sec from an initial height of 200 ft above the ground is given by the
\n" ); document.write( "following formula:
\n" ); document.write( "s(t) = −16t^2 + 96t + 200\r
\n" ); document.write( "\n" ); document.write( "(a) At what moment in time is the height of the projectile the greatest?
\n" ); document.write( "(b) At what moment in time does the projectile return to the ground?
\n" ); document.write( "(c) The slope of the line through points (t0, s(t0)) and (t1, s(t1)) is the average velocity
\n" ); document.write( "on the time interval [t0, t1]. What is the average velocity of the particle from the
\n" ); document.write( "moment it was launched to the moment when it reaches its greatest height?
\n" ); document.write( "

Algebra.Com's Answer #651560 by Alan3354(69443)\"\" \"About 
You can put this solution on YOUR website!
The height of a projectile t seconds after being launched with an initial
\n" ); document.write( "velocity of 96 ft/sec from an initial height of 200 ft above the ground is given by the
\n" ); document.write( "following formula:
\n" ); document.write( "s(t) = -16t^2 + 96t + 200\r
\n" ); document.write( "\n" ); document.write( "(a) At what moment in time is the height of the projectile the greatest?
\n" ); document.write( "\"moment in time\" is redundant.
\n" ); document.write( "---
\n" ); document.write( "(a) At what time is the height of the projectile the greatest?
\n" ); document.write( "It's the vertex of the parabola, at t = -b/2a
\n" ); document.write( "t = -96/-32 = 3 seconds.
\n" ); document.write( "=========================
\n" ); document.write( "(b) At what time does the projectile return to the ground?
\n" ); document.write( "When s(t) = 0.
\n" ); document.write( "-16t^2 + 96t + 200 = 0
\n" ); document.write( "-2t^2 + 12t + 25 = 0
\n" ); document.write( "\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( "
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation \"ax%5E2%2Bbx%2Bc=0\" (in our case \"-2x%5E2%2B12x%2B25+=+0\") has the following solutons:
\n" ); document.write( "
\n" ); document.write( " \"x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca\"
\n" ); document.write( "
\n" ); document.write( " For these solutions to exist, the discriminant \"b%5E2-4ac\" should not be a negative number.
\n" ); document.write( "
\n" ); document.write( " First, we need to compute the discriminant \"b%5E2-4ac\": \"b%5E2-4ac=%2812%29%5E2-4%2A-2%2A25=344\".
\n" ); document.write( "
\n" ); document.write( " Discriminant d=344 is greater than zero. That means that there are two solutions: \"+x%5B12%5D+=+%28-12%2B-sqrt%28+344+%29%29%2F2%5Ca\".
\n" ); document.write( "
\n" ); document.write( " \"x%5B1%5D+=+%28-%2812%29%2Bsqrt%28+344+%29%29%2F2%5C-2+=+-1.63680924774785\"
\n" ); document.write( " \"x%5B2%5D+=+%28-%2812%29-sqrt%28+344+%29%29%2F2%5C-2+=+7.63680924774785\"
\n" ); document.write( "
\n" ); document.write( " Quadratic expression \"-2x%5E2%2B12x%2B25\" can be factored:
\n" ); document.write( " \"-2x%5E2%2B12x%2B25+=+%28x--1.63680924774785%29%2A%28x-7.63680924774785%29\"
\n" ); document.write( " Again, the answer is: -1.63680924774785, 7.63680924774785.\n" ); document.write( "Here's your graph:
\n" ); document.write( "\"graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+-2%2Ax%5E2%2B12%2Ax%2B25+%29\"

\n" ); document.write( "\n" ); document.write( "======
\n" ); document.write( "Ignore the negative solution.
\n" ); document.write( "t =~ 7.637 seconds
\n" ); document.write( "===============
\n" ); document.write( "(c) The slope of the line through points (t0, s(t0)) and (t1, s(t1)) is the average velocity
\n" ); document.write( "on the time interval [t0, t1]. What is the average velocity of the particle from the
\n" ); document.write( "moment it was launched to the moment when it reaches its greatest height?
\n" ); document.write( "--
\n" ); document.write( "t0 = 0, t1 = 3
\n" ); document.write( "s0 = 200, s1 = s(3) = -16*9 + 96*3 + 200 = 344
\n" ); document.write( "--> (344-200)/3 = 48 ft/sec
\n" ); document.write( "
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