Answer:
Error in the equation, but see below for the process.
Explanation:
p(t)=(400t+80,200t−100,41t−16t2
What is p? The problem says "path," but does that include the x, y, and z axes? Also, is the formula written correctly? I see two entrees of Number*t, and the last term, -16t2 also seems odd.
For this answer, we'll assume p(t) is the elevation (height) of the cannonball in feet at time x. I'll also assume the the final t is t^2, not t2.
p(t)=400t+80,200t−100,41t−16t^2
p(t)= t*(400 + 80,200 + 100,410)−16t^2
p(t) = 181010t - 16t^2
This seems an extremely odd equation, so please check.
We assume it is launched at 0 feet, ground level. We can solve the equation for p(t)=0. We know one solution will be at time 0. But solving the quadratic equation will give us the second time for which p(t) = 0.
p(t) = 181010t - 16t^2
0 = 181010t - 16t^2The two roots are t = 0 seconds (at launch) and t = (90505/8) or 11213 seconds
We can also graph the line and look for the intersection with the x axis. See the attached graph. The two points the line intersects the x axis are t = o and t = 11400 seconds.
Please check the equation, correct it, and repeat these steps.