A basketball trajectory at a different height

Step 1: Figure out the knowns and unknowns

The rim is at x = 6.5 m and y = 3.0 m relative to a release point at (0, 3.5) m. We can evaluate the trajectory height at x = 6.5 m, or find the horizontal coordinate where the descending ball reaches y = 3.0 m.

What we do have is an initial point for $x_0$, which we can say is $0\text{m}$. You could also say that the shot starts at $x_0=6.5\text{m}$, it’s up to you. We also can use time as a second unknown.

Step 2: Set up the system of equations.

For this problem, we use the same equation as

in Example Problem 3.2.3.

$x=x_0+{\overrightarrow{v}}_{\mathrm{0,}x}\mathrm{\Delta }t+\frac{1}{2}{\overrightarrow{a}}_x{\left(\mathrm{\Delta }t\right)}^2$ for the $x$-direction

$y_y=y_0+{\overrightarrow{v}}_{\mathrm{0,}y}\mathrm{\Delta }t+\frac{1}{2}{\overrightarrow{a}}_y{\left(\mathrm{\Delta }t\right)}^2$ for the $y$-direction​
The system we get is:

x = (10 cos 25°)t = (9.06308 m/s)t.

y = 3.5 m + (10 sin 25°)t − ½(9.81 m/s²)t² = 3.5 m + (4.22618 m/s)t − (4.905 m/s²)t².

At the rim’s height, 0 = 0.5 + x tan 25° − [9.81/(2·10² cos²25°)]x², with x in metres.

Solving this equation gives us:

The descending ball reaches y = 3.0 m at x ≈ 8.764 m, so it overshoots that rim-height crossing by 2.264 m. At the actual rim location x = 6.5 m, its center is approximately 4.008 m high, 1.008 m above the rim. Under the point-projectile model, the shot misses.