My First Problem about Physics

The highest the basketball player can throw the ball will be if he threw it at a 90 degree angle (straight up) With an initial velocity of 5 m/s, the highest the ball will go is 1.27 meters above his initial height of 2 m. Therefore, the maximum height of the ball, above ground, is 3.27 meters, which is lower than the hoop height of 5 meters. So, the answer is no.

Let's identify the given:

I used where the basketball player is standing as the reference point (origin).

$\textbf{r}_i = 2 \hat{j} \\ \textbf{r}_f = 6 \hat{i} + 5 \hat{j} \\ |\textbf{V}| = 5 \ \text{m/s} \\ \textbf{a} = -9.8 \hat{j}$

Then, we will use this 2 formulas:

$\textbf{V} = V_{ox}\hat{i} + V_{oy} \hat{j}$

and

$\textbf{r}_f = \textbf{r}_i + \textbf{V}t + 0.5\textbf{a} t^2$

---

$\begin{aligned} \textbf{V} &= V_{ox}\hat{i} + V_{oy} \hat{j} \\ &= 5\cos\theta \hat{i} + 5\sin\theta \hat{j} \end{aligned}$

---

$\begin{aligned} \textbf{r}_f &= \textbf{r}_i + \textbf{V}t + 0.5\textbf{a} t^2 \\ \implies 6 \hat{i} + 5 \hat{j} &= 2 \hat{j} + \left( 5\cos\theta \hat{i} + 5\sin\theta \hat{j}\right)t - 4.9t^2 \hat{j} \end{aligned}$

By analogy,

$6 = 5t\cos \theta \implies t = \dfrac{6}{5\cos \theta }$

and

$\begin{aligned} 5 &= 2 + 5t\sin\theta - 4.9t^2 \\ 3 &= 5\left(\dfrac{6}{5\cos \theta }\right)(\sin \theta) - 4.9 \left( \dfrac{6}{5\cos \theta } \right)^2 \\ 75\cos ^2 \theta &= 150 \sin\theta\cos\theta - 4.9(36) \\ 75\cos ^2 \theta &= 150\left( \sqrt{1 - \cos ^2 \theta} \right) \cos\theta - 4.9(36) \\ \left( 75\cos ^2 \theta + 4.9(36)\right)^2 &= 150^2 (1 - \cos ^2 \theta)(\cos ^2 \theta) \\ 28125 a^2 + 3960 a + 31117 &= 0 \hspace{15mm} \color{#3D99F6}{a = \cos ^2 \theta \ ; \ a \le 1} \end{aligned}$

If you get the discriminant of that quadratic equation, you'll see that $a$ is imaginary, hence, there is $\textbf{NO}$ such $\theta$ applicable.