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.
Quadratic Model: -16t2+192t+32
1. How high does the cannonball go? 608 ft (Remember you are looking for a specific part of the vertex.)
2. How long is the cannonball in the air? 12 seconds (Remember you can use the quadratic formula.)
Steps to Solve:
1. Plug the numbers from the situation above into the quadratic model representing an object thrown into the air.
2. Then plug the numbers into the function that finds the X of the vertex of the parabola (-b/2a).
3. When the X of the vertex is obtained plug that number into the quadratic model and that will be how high the cannon goes.
4. Plug the quadratic model into the calculator and go to the table. The highest number under the X column (before the numbers start to turn negative) will be how long the cannonball stayed in the air.
I got the same answer, that's a relief :) I like how the steps are layed out, and how you use the calculator's ability to its full extent!! (yeah I sound like a nerd) Very simple to follow
ReplyDeleteI like how you showed all the steps clearly. Luckily, I got the same answer! You explained it well.
ReplyDeleteThe steps for this were so clear and THANK GOODNESS I got the same answer! Easy-peasy to read and understand. AWESOME!
ReplyDelete