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,
![x=v\cos\theta t](https://tex.z-dn.net/?f=x%3Dv%5Ccos%5Ctheta%20t)
On substituting the known values,
![\begin{gathered} x=20\times\cos35\degree\times t \\ \implies x=16.4t \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20x%3D20%5Ctimes%5Ccos35%5Cdegree%5Ctimes%20t%20%5C%5C%20%5Cimplies%20x%3D16.4t%20%5Cend%7Bgathered%7D)
The change in the height, y of the basketball is given by,
![y=-v\sin\theta t+\frac{1}{2}gt^2](https://tex.z-dn.net/?f=y%3D-v%5Csin%5Ctheta%20t%2B%5Cfrac%7B1%7D%7B2%7Dgt%5E2)
Where g is the acceleration due to gravity.
On substituting the known values,
![\begin{gathered} y=-20\times\sin35\degree\times t+\frac{1}{2}\times32\times t^2 \\ \implies y=-11.5t+16t^2 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20y%3D-20%5Ctimes%5Csin35%5Cdegree%5Ctimes%20t%2B%5Cfrac%7B1%7D%7B2%7D%5Ctimes32%5Ctimes%20t%5E2%20%5C%5C%20%5Cimplies%20y%3D-11.5t%2B16t%5E2%20%5Cend%7Bgathered%7D)
Final answer:
The parametric equations describing the shot are