Answer to Question #134848 in Physics for Charlie

Question

A garden hose pointed at an angle of 25 drgrees above the horizontal splashes water on a sunbather lying on the ground 4.4m away in the horizontal direction. If the hose is held 1.4m above the ground at what speed does the water leave the nozzel

Answer 1

Write the expression for range:

$$R=v\text{ cos}\theta\cdot t.$$

The time is

$$t=t_{up}+t_{down}.\\\space\\ t_{up}=\frac{v\text{ sin}\theta}{g},\\\space\\ t_{down}=\sqrt{\frac{2H}{g}}.$$

The height the water will fall down is

$$H=\frac{gt_{up}^2}{2}+h=\frac{v^2\text{sin}^2\theta}{2g}+h.$$

Thus, the time down becomes

$$t_{down}=\sqrt{\frac{2\big(\frac{v^2\text{sin}^2\theta}{2g}+h\big)}{g}}.$$

Now, express the time from the range (see the first equation):

$$t=\frac{R}{v\text{ cos}\theta}=t_{up}+t_{down},\\\space\\ \frac{R}{v\text{ cos}\theta}=\frac{v\text{ sin}\theta}{g}+\sqrt{\frac{2\big(\frac{v^2\text{sin}^2\theta}{2g}+h\big)}{g}}.$$

All variables are given except the initial velocity. So, we can easily solve this equation:

$$v=5.8\text{ m/s}.$$