Question: A cannonball is shot upward from the upper deck of a fort with an initial velocity of 192 feet per second. The deck is 32 feet above the ground.
How to Solve:
1. Read the problem and plug in the velocity and initial height into the equation.
2. Plug the revised equation into to y= key on your graphing calculator.
3. Scroll through to numbers until you see the highest possible Y value
4. When you find the Y value, that will be how high the cannonball goes.
5. The corresponding X value will be how long the cannonball is in the air.
Quadratic model: -16t2+(192)t+(32)
How high does the cannonball go? ~ 608 feet
How long is the cannonball in the air? 6 seconds
Chemical Balance
C2H5OH + O2 => CO2 + H2O We then find coeifficiants to balance the unbalanced equation.
C2H5OH + 3 O2 => 2 CO2 + 3 H2O
