Answer
$\large{y = -\frac{3}{320} (x-80)^2+60 \,\,\,\,for\,\,0\leq x \leq160}$
Work Step by Step
The trajectory of the human cannonball is assumed to be that of a parabola.
We consider the opening of the cannon as the origin of the coordinate system $(0,0)$.
The net that catches the human cannonball is on the same vertical level and 160 ft away, thus it's coordinates are $(160,0)$.
The vertex of the parabola lies at $(80,60)$.
The equation of the parabola is:
$$y = a(x-h)^2+k$$
$\because$ The vertex of the parabola $(h,k)$ is known to be at $(80,60)$
$$\therefore y=a(x-80)^2+60$$
To find $a$, we substitute the landing point $(160,0)$ in the equation.
$$0 = a(160-80)^2+60$$
$$-60 = a(80)^2$$
$$a = -\frac{60}{6400} = -\frac{3}{320}$$
$$\therefore y = -\frac{3}{320} (x-80)^2+60 \,\,\,\,for\,\,0\leq x \leq160$$
$\large{Verification}$
$$y(0) = -\frac{3}{320} (0-80)^2+60=-\frac{3}{320} (6400)+60=-60+60 = 0$$
$$y(160) = -\frac{3}{320} (160-80)^2+60= -\frac{3}{320} (80)^2+60=-\frac{3}{320} (6400)+60=-60+60 = 0$$
