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,

On substituting the known values,

The change in the height, y of the basketball is given by,

Where g is the acceleration due to gravity.
On substituting the known values,

Final answer:
The parametric equations describing the shot are