Question

A basketball player shoots a free throw, where the position of the ball is modeled by x = (26cos 50°)t and y = 5.8 + (26sin 50°)t − 16t 2. What is the height of the ball, in feet, when it is 13 feet from the free throw line? Round to three decimal places.

Answer 1

Given:

The position of the ball is modeled by:

$\begin{gathered} x\text{ = \lparen26cos50\rparen t} \\ y\text{ = 5.8 + \lparen26sin50\rparen t - 16t}^2 \end{gathered}$

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

$\begin{gathered} 13\text{ = \lparen26cos50\rparen t} \\ t\text{ = }\frac{13}{26cos50} \\ =\text{ 0.7779 sec} \end{gathered}$

Substituting the value of t into the equation for y to get the height:

$\begin{gathered} y\text{ = 5.8 + \lparen26sin50\rparen}\times\text{ 0.7779 - 16 }\times\text{ \lparen0.7779\rparen}^2 \\ =\text{ 11.611} \end{gathered}$

Hence, the height of the ball is 11.611 feet