Physics · Mechanics · Grade 10

Basketball: the free throw

At what angle must the ball be thrown to drop into the basket?

The situation

1 second left in the game, and the score is tied. You are standing on the free-throw line — the basket is 4.2 m away and 3.05 m high. At what angle and with what speed must you throw the ball for it to go through the hoop? And what about on the Moon or on Jupiter?

Topic

Projectile launched at an angle. Basketball: free throw and three-point shot

Key ideas

Projectile launched at an angle
Velocity components
Air resistance

Formulas

x = v_{0} \cos α \cdot t, y = h + v_{0} \sin α \cdot t - \frac{g t^{2}}{2}

Two independent motions: horizontal — uniform, vertical — uniformly accelerated.

y (x) = h + x tg α - \frac{g x^{2}}{2 v_{0}^{2} \cos^{2} α}

The equation of the trajectory — a parabola.

v_{0} = \sqrt{\frac{g d^{2}}{2 \cos^{2} α (d tg α - Δ y)}}

The speed needed to hit the centre of the hoop ( $d$ — distance, $Δ y$ — how much higher the hoop is than the hand).

T = \frac{2 v_{0} \sin α}{g}, L = \frac{v_{0}^{2} \sin 2 α}{g}, H = \frac{v_{0}^{2} \sin^{2} α}{2 g}

When the ball returns to the same height: time of flight, range, maximum height.

a_{τ} = - g \frac{v_{y}}{v}, a_{n} = g \frac{v_{x}}{v}, R = \frac{v^{2}}{a_{n}}

Tangential and normal acceleration, radius of curvature of the trajectory (no air).

Theory

Two motions at once

Without air resistance, the only force on the ball is its weight, and it points downwards. So the horizontal velocity $v_{x}$ stays the same throughout the flight, while the vertical velocity $v_{y}$ decreases by $g$ every second. On the graph, $v_{x}$ is a horizontal line and $v_{y}$ is a sloping straight line.

Why not 45°?

A target at the same height is reached with the smallest speed at 45°. But the hoop is about 1 m above the hand, and the ball must also enter the hoop from above: the steeper it comes down, the larger the “visible” opening of the hoop. That is why a free throw is usually shot at more than 50°.

Air resistance

A basketball is light and large, so at 7 m/s the drag force $F_{d} = \frac{1}{2} ρ C_{d} A v^{2}$ is about 10 % of the ball's weight. It slows the ball down: $v_{x}$ decreases slowly, and towards the end of the flight the parabola becomes steeper.

Curvilinear motion: normal and tangential acceleration

Let us split the acceleration $\vec{g}$ into two components: along the velocity — the tangential one, $a_{τ}$ (it changes the magnitude of the velocity), and perpendicular to it — the normal one, $a_{n}$ (it turns the direction of the velocity). On the way up $a_{τ} < 0$ — the ball slows down; on the way down $a_{τ} > 0$ . At the highest point $a_{τ} = 0$ , $a_{n} = g$ , and the radius of curvature is smallest: $R = v_{x}^{2} / g$ . Watch the green arrows in the scene and the values of $a_{τ}$ , $a_{n}$ , $R$ in the “Measurements” section. Coordinate graph: $x (t)$ is a straight line, $y (t)$ is a parabola; eliminating $t$ from them gives the equation of the trajectory $y (x)$ .

Lab: the launcher

A spring launcher fires a small ball from height $h$ , and the spot where it lands on the floor is measured with a tape measure. Horizontal launch ( $α = 0$ ): $t = \sqrt{2 h / g}$ , $L = v_{0} t$ , so $v_{0} = L \sqrt{g / (2 h)}$ ; the final speed is $v = \sqrt{v_{0}^{2} + 2 g h}$ , and the angle of impact is $tg β = \sqrt{2 g h} / v_{0}$ . At an angle: the horizontal velocity does not change, so from the time of flight $T$ and the range $L$ we get $v_{0} = \frac{L}{T \cos α}$ , $h_{m a x} = h + \frac{v_{0}^{2} \sin^{2} α}{2 g}$ ; if the ball returns to the same level, $L = \frac{v_{0}^{2} \sin 2 α}{g}$ and $h_{m a x} = \frac{L}{4} tg α$ . $α = 90 °$ — a vertical launch upwards, $v_{0} = 0$ — free fall (in detail — in the “Free fall and air” experiment). The ball is small and heavy, so air resistance is neglected.

On other planets

The required speed is $v_{0} \propto \sqrt{g}$ . On the Moon $g$ is 6 times smaller, so at the same angle the speed needed is $\sqrt{6} \approx 2.4$ times smaller, and the flight lasts that many times longer.

Did you know?

Studies show that the best launch angle for a free throw is about 50–55° — at this angle a small error in speed affects the result the least.
The hoop is 3.05 m (10 ft) above the floor. This height has not changed since 1891, when James Naismith nailed peach baskets to the balcony of a gym.

Inside the experiment

The experiment starts from a real-life situation: you predict the result, change the controls, run the experiment and draw a conclusion. Inside are missions, theory, a quiz and a guided tour in which the camera moves and explains the key points.

More experiments in this section

Circular motion: the bicycle — Grade 7
Pascal's law and hydraulics — Grade 7
Elasticity and deformation — Grade 9
Why doesn't a ship sink? — Grade 10
Simple machines: lever, pulley — Grade 10
Momentum and its conservation — Grade 10
Banked turn — Grade 10
Centre of gravity and equilibrium — Grade 10
Aeroplane wing — Grade 10
Dynamics of a rigid body — Grade 10