When the equation of it’s pathway can be modeled by h = -16r(-16r) + 128r Find the maximum height “h” feet of the cannonball. Find the “t” seconds it will take for the cannonball to reach the group.

Question

It’s a parabola

Answer 1

I guess you mean -16t²+128t for the height. We can write this as -16(t²-8t). We can complete the square: -16(t²-8t+16-16) = -16(t-4)²+256. When t=4 this expression for the height is maximum at 256 feet, because there is nothing to subtract from 256. This means that maximum height is reached at 4 seconds.

This the vertex of the parabola and is halfway on the trajectory, so at t=8, h=0, when it reaches the ground. You can also see this from the original equation h=-16t²+128t=16t(-t+8) when t=8, h=0. Of course, when t=0, h=0 at the start of the trajectory.