Given:
The angle of projection of the basketball, θ=35°
The height at which the ball leaves the hand, h=7 ft
The initial velocity of the basketball, v=20 ft/s
To find:
The parametric equations describing the shot.
Explanation:
The range, x of the basketball is given by,
[tex]x=v\cos\theta t[/tex]
On substituting the known values,
[tex]\begin{gathered} x=20\times\cos35\degree\times t \\ \implies x=16.4t \end{gathered}[/tex]
The change in the height, y of the basketball is given by,
[tex]y=-v\sin\theta t+\frac{1}{2}gt^2[/tex]
Where g is the acceleration due to gravity.
On substituting the known values,
[tex]\begin{gathered} y=-20\times\sin35\degree\times t+\frac{1}{2}\times32\times t^2 \\ \implies y=-11.5t+16t^2 \end{gathered}[/tex]
Final answer:
The parametric equations describing the shot are
[tex]\begin{gathered} \begin{equation*} x=16.4t \end{equation*} \\ \begin{equation*} y=-11.5t+16t^2 \end{equation*} \end{gathered}[/tex]
