Projectile Motion Gets the Hose

Abstract

Students take a weekly quiz in our introductory physics course. During the week in which material focused on projectile motion, we not-so-subtly suggested what problem the students would see on the quiz. The quiz problem was an almost exact replica of a homework problem1 we worked through in the class preceding the quiz. The goal of the problem is to find the launch speed if the final horizontal and vertical positions and launch angle are given. Figure 1 shows a schematic of the trajectory.

Liquid jet leaping from a free surface

“…1 Liquid jets fall into two basic categories, 2 submerged jets and non-submerged jets, depending on whether the surrounding medium is the same as the medium of the jet. In the case of a typical non-submerged jet, [3][4][5][6] the liquid jet is ejected from a nozzle above the liquid surface into the air, and the trajectory, except for vertical ejection, can be roughly described by the parabolic curve y = x v 0 cosθ v 0 sinθ − 1 2 g x v 0 cosθ , where y is the vertical displacement, x is the horizontal displacement, g is acceleration due to gravity, θ is the slant angle relative to the horizon, and v 0 is the average velocity of the liquid jet from the nozzle. In the case of a typical submerged jet, 1 the liquid jet is ejected from a nozzle below the liquid surface and does not leap out of the free surface.…”