Question

 A fountain of water is located at the center of a circular pool. Not wishing 

to get his feet wet, a student walks around the pool and measures its 

circumference to be 16.5 m. Next, the student stands at the edge of the pool 

and uses a protractor to gauge the angle of elevation at the bottom of the 

fountain to be 55.0°. How high is the fountain? 

Answer 1

The water passes these R=16.5 meters in

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

On the other hand, a half of this this time is required to reach the highest point from the bottom:

$$\frac t2=\frac{v_y}{g}=\frac{v\text{ sin}\theta}{g},$$

substitute the time from the first equation:

$$\frac{R}{v\text{ cos}\theta}=\frac{2v\text{ sin}\theta}{g},\\\space\\ v^2=\frac{Rg}{\text{sin}(2\theta)}.$$

During a half of this time, on the other hand, the water raises up, or makes the first part of the trajectory from the bottom to the highest point:

$$h=\frac{v_y^2}{2g}=\frac{(v\text{ sin}\theta)^2}{2g}= \frac{\big[\frac{Rg}{\text{sin}(2\theta)}\big]\text{sin}^2\theta}{2g}=\\\space\\ =\frac R4\text{ tan}\theta=5.89\text{ m}.$$