Question

The regulation height of a basketball hoop is 10 feet. Let the location of thebasket be represented in the coordinate plane by the point (0, 10). Let the ballbe thrown at a 45° angle with the ground.1. Suppose Nancy is standing a horizontal distance of 10 feet from thebasket at the point (-10, 0), and she shoots a basket from 6 feet in theair with an initial velocity of 22 ft/s.a. Write parametric equations that represent the ball's motion throughthe air.b. Graph the parametric equations on your calculator in an appropriatewindow and sketch the results below.

Answer 1

SOLUTION:

a. The parametric equations that represent the balls motion is;

$\begin{gathered} x(t)=x_0+(v_0cos\theta)t \\ y(t)=y_0+(v_0sin\theta)t+0.5gt^2 \end{gathered}$

Inserting the values;

$\begin{gathered} x(t)=-10+(22cos45)t \\ y(t)=6+(22sin45)t+0.5(-32)t^2 \end{gathered}$

Simplifying, we have;

$\begin{gathered} x(t)=15.56t-10 \\ y(t)=-16t^2+15.56t+6 \end{gathered}$

b. The graph of the parametric equation is given below;