We are given that a dolphin moves describing a projectile motion. This can be represented in the following graph of position vs time:
Since the dolphin moves horizontally as he goes through the hoop this means that the hoop is at the maximum height of the motion. The maximum height of a projectile motion is given by:
[tex]h_{\max }=\frac{v^2\sin ^2\theta}{2g}[/tex]
Where:
[tex]\begin{gathered} h_{\max }=\text{ max}imum\text{ height} \\ v=velocity_{} \\ \theta=\text{ initial angle} \\ g=\text{ acceleration of gravity} \end{gathered}[/tex]
Now, we plug in the values:
[tex]h_{\text{max}}=\frac{(10\frac{m}{s})^2(\sin (41))^2}{2(9.8\frac{m}{s^2})}[/tex]
Solving the operations:
[tex]h_{\max }=2.2m[/tex]
Therefore, the hoop is at 2.2 meters above the water.
