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: h = -16t2+192t+32
First, to find the time, plug the data above into the quadratic equation, which is
where (-16) is "a"when is (1/2) of Earth's gravity which is (32 f/s^2), (192 f/s) is "b", and (32 ft) is "c". When you calulate and simplify the equation, you get (-2) and (12.2), and you can't have negative time, so you chose the posetive answer, (12.2 seconds).
After this, you plug the original info into another equation,x = -b/2a, which gives you the axis of symmetry. once you calculate and simplify it, you find the answer of (6).
You pulg (6) into the original quadratic model.
Once you calculate and simplify it, you get the vertex of (6, 608), which means after (6) seconds, the cannon ball reachs its highest point of motion at (608 ft).
First, to find the time, plug the data above into the quadratic equation, which is
where (-16) is "a"when is (1/2) of Earth's gravity which is (32 f/s^2), (192 f/s) is "b", and (32 ft) is "c". When you calulate and simplify the equation, you get (-2) and (12.2), and you can't have negative time, so you chose the posetive answer, (12.2 seconds). After this, you plug the original info into another equation,x = -b/2a, which gives you the axis of symmetry. once you calculate and simplify it, you find the answer of (6).
You pulg (6) into the original quadratic model.
Once you calculate and simplify it, you get the vertex of (6, 608), which means after (6) seconds, the cannon ball reachs its highest point of motion at (608 ft).
1. How high does the cannonball go? (608 feet)
2. How long is the cannonball in the air? (12.2 seconds)
2. How long is the cannonball in the air? (12.2 seconds)
