Basketball Trajectory: Will It Score?

This is real basketball physics: a player releases the ball from height $h_0$ at speed $v_0$ and angle $\theta$, while the hoop sits at height $h_{\text{rim}}$ and horizontal distance $d$ away. Does the trajectory pass through the rim? We solve it the way a physicist (or a serious shooter) would: find the ball's height at the rim's horizontal position, then compare.

The Physics

With the release point as the origin (so the rim is at $(d,\, h_{\text{rim}} - h_0)$ in our local coordinates), the parametric equations are:

$x(t) = v_0 \cos\theta\, t \qquad y(t) = v_0 \sin\theta\, t - \tfrac{1}{2}\,g\,t^{2}$

We can eliminate $t$ to get $y$ directly as a function of $x$ — this is the trajectory equation:

$y(x) = x\tan\theta - \dfrac{g\,x^{2}}{2\,v_0^{2}\,\cos^{2}\theta}$

To check whether the ball clears the rim, plug $x = d$ into this equation and add the release height. If the result is close to $h_{\text{rim}}$, it's a basket.

Step-by-Step Solution

Given:

• Release velocity: $v_0 = 10\;\text{m/s}$
• Release angle: $\theta = 60°$
• Release height: $h_0 = 2\;\text{m}$
• Horizontal distance to rim: $d = 5\;\text{m}$
• Rim height: $h_{\text{rim}} = 3.0\;\text{m}$
• Gravity: $g = 9.81\;\text{m/s}^{2}$

Find: Does the ball pass through the rim?

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Step 1 — Decompose the launch velocity.

$v_{0x} = 10\cos 60° = 10 \times 0.5 = 5.000\;\text{m/s}$

$v_{0y} = 10\sin 60° = 10 \times \dfrac{\sqrt{3}}{2} \approx 8.660\;\text{m/s}$

Step 2 — Find the time at which $x = d$.

Since horizontal velocity is constant:

$t = \dfrac{d}{v_{0x}} = \dfrac{5}{5} = 1.000\;\text{s}$

Step 3 — Find the vertical position at that time (relative to release).

$y_{\text{rel}}(t) = v_{0y}\,t - \tfrac{1}{2}\,g\,t^{2}$

$y_{\text{rel}}(1) = (8.660)(1) - \tfrac{1}{2}(9.81)(1)^{2}$

$y_{\text{rel}}(1) = 8.660 - 4.905 \approx 3.755\;\text{m}$

Step 4 — Add the release height to get height above the floor.

$y_{\text{floor}} = y_{\text{rel}} + h_0 = 3.755 + 2.000 \approx 5.755\;\text{m}$

Step 5 — Compare to the rim height.

$\Delta = y_{\text{floor}} - h_{\text{rim}} = 5.755 - 3.000 \approx 2.755\;\text{m}$

The ball is 2.76 m above the rim when it crosses $x = 5\;\text{m}$ — that's almost three meters too high. The trajectory sails right over the basket.

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Answer: The shot misses (way too high). The ball is at $\approx 5.76\;\text{m}$ above the floor when it reaches the rim's horizontal distance, which is 2.76 m above the 3 m rim. To score from this distance with this release height and angle, you'd need to lower the angle to about 41° (so the trajectory comes down faster) or reduce the release speed to about 7.7 m/s.

Try It

• Lower the angle to about 41° — see the ball drop right into the rim.
• Adjust release velocity — try 7.7 m/s at 60° as an alternative scoring solution.
• Move the rim distance — closer shots forgive a higher arc, farther shots need flatter trajectories.
• The green ✓ SWISH! indicator lights up when the trajectory clears the rim within ±0.2 m tolerance.