Water squirting out of a horizontal nozzle held $4$ ft above the ground describes a parabolic curve with the vertex at the nozzle. If the stream of water drops $1$ ft in the first $10$ ft of horizontal motion, at what horizontal distance from the nozzle will it strike the ground?
So far I have visualized that this parabola is probably a $y^2=4px$ or $y^2 = x$, the focus must be some $(a,4)$ and its directrix some $(-a,4)$, with the vertex of this parabola at $y=4.$ The $y$ axis should be describing the height and the $x$ axis horizontal distance.
For it to hit the ground I should be finding some $x$ value when $y^2 = 0.$ However just by simply substituting it to be $0$ causes the equation to have no $x$ value.
I have kinda found out the $p$ value of this parabola by doing $\sqrt{10^2+(3-p)^2} = (3+p)^2$ where $p = 25/3.$ (I used $3$ as the question says it drops by $1$ ft), but it doesn't seem to help as described above.
I wonder if I should be trying something else but I am stuck.