Answer to Question #131724 in Physics for Devin Maynard

Question

. Water from an irrigation sprinkler lands on a corn plant 37 meters away. If the sprinkler angle is 40˚,
with what initial velocity was the water launched?

Answer 1

Initially, water velocity had two components, y and x. The range (37 m) is defined by both:

$$v_y=v\text{ sin}\theta,\\ v_x=v\text{ cos}\theta.$$

The time required for a mass of water to go up:

$$t_\text{up}=\frac{v_y}{g}.$$

It is easy to understand that the time for the water to reach the ground level again is

$$t=2t_\text{up}=2\frac{v_y}{g}=\frac{2v\text{ sin}\theta}{g}.$$

The range is determined by the horizontal component only:

$$R=v_xt=v\text{ cos}\theta\cdot\frac{2v\text{ sin}\theta}{g}=\frac{v^2\text{sin}(2\theta)}{g}.$$

Express and calculate the initial velocity:

$$v=\sqrt{\frac{gR}{\text{sin}(2\theta)}}=19.2\text{ m/s}.$$