The three-point shot

Problem

A basketball player takes a three-point shot. The ball leaves from a height of $h_0 = 2{,}1\;\text{m}$ at a horizontal distance of $6{,}75\;\text{m}$ from the hoop, which is at a height of $3{,}05\;\text{m}$. The acceleration is $\vec{a} = \begin{pmatrix} 0 \\ -9{,}81 \end{pmatrix}\;\text{m/s}^2$. Determine the required initial speed, knowing that the ball is thrown at an angle of $55^\circ$ to the horizontal.

Solution

Setup. We choose the origin at the launch point and the axes $x$ (horizontal) and $y$ (vertical, upwards). The parabolic motion splits into two independent motions: uniform along $x$, uniformly accelerated along $y$ with $a_y = -9{,}81\;\text{m/s}^2$.

Let us resolve the initial speed using trigonometry:

Event A (launch): $t_A = 0$, $\vec{S}_A = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$, $\vec{v}_A = \begin{pmatrix} v_0\cos 55^\circ \\ v_0\sin 55^\circ \end{pmatrix}$

Event B (hoop): $\vec{S}_B = \begin{pmatrix} 6{,}75 \\ 0{,}95 \end{pmatrix}$, where the height is the height difference $3{,}05 - 2{,}10 = 0{,}95\;\text{m}$.

The parametric equations of motion between A and B are:

$\begin{cases} 6{,}75 = v_0\cos 55^\circ\cdot t_B \\ 0{,}95 = v_0\sin 55^\circ\cdot t_B - \tfrac{1}{2}\cdot 9{,}81\cdot t_B^2 \end{cases}$

Eliminating time. From the first equation we obtain the instant of arrival at the hoop:

$t_B = \frac{6{,}75}{v_0\cos 55^\circ}$

Substituting into the second equation. Note that in the first term $v_0\sin 55^\circ\cdot t_B = 6{,}75\cdot\tan 55^\circ$, while the second term contains $t_B^2$:

$0{,}95 = 6{,}75\,\tan 55^\circ - \frac{9{,}81}{2}\left(\frac{6{,}75}{v_0\cos 55^\circ}\right)^2$

Solving. With $\tan 55^\circ \approx 1{,}428$ and $\cos 55^\circ \approx 0{,}5736$, let us work out the numerical terms:

$0{,}95 = 9{,}64 - 4{,}905\cdot\frac{45{,}56}{0{,}3290}\cdot\frac{1}{v_0^2} = 9{,}64 - \frac{679{,}3}{v_0^2}$

Isolating the term with $v_0^2$ and solving:

$\frac{679{,}3}{v_0^2} = 9{,}64 - 0{,}95 = 8{,}69 \quad\Rightarrow\quad v_0^2 = \frac{679{,}3}{8{,}69} \approx 78{,}2 \quad\Rightarrow\quad v_0 = \sqrt{78{,}2} \approx 8{,}84\;\text{m/s}$

$\ev{v_0 \approx 8{,}84\;\mathrm{m/s}}$

Links

Topics: Cinematica Concepts: Moto parabolico Skills: Scomposizione vettoriale · Uso della trigonometria Methods: Scomposizione in componenti cartesiane