Given:
The position of the ball is modeled by:
[tex]\begin{gathered} x\text{ = \lparen26cos50\rparen t} \\ y\text{ = 5.8 + \lparen26sin50\rparen t - 16t}^2 \end{gathered}[/tex]
The ball is 13 feet from the free throw line.
This implies that the horizontal distance (x) that the ball has traveled is 13 feet
Substituting the value of x into the formula and solving for t
[tex]\begin{gathered} 13\text{ = \lparen26cos50\rparen t} \\ t\text{ = }\frac{13}{26cos50} \\ =\text{ 0.7779 sec} \end{gathered}[/tex]
Substituting the value of t into the equation for y to get the height:
[tex]\begin{gathered} y\text{ = 5.8 + \lparen26sin50\rparen}\times\text{ 0.7779 - 16 }\times\text{ \lparen0.7779\rparen}^2 \\ =\text{ 11.611} \end{gathered}[/tex]
Hence, the height of the ball is 11.611 feet
