To solve for the initial speed $v_0$ of the basketball, we can break the problem down into two parts: horizontal and vertical motion. Since the motion is two-dimensional, we'll apply the principles of projectile motion.

Step 1: Horizontal Motion

The horizontal distance $x$ to the hoop is given by:

$x = v_0 \cos(\alpha) \cdot t$

Where:

• $v_0$ is the initial speed of the ball.
• $\alpha$ is the angle of release.
• $t$ is the time of flight.

We can solve for $t$ in terms of $v_0$:

$t = \frac{x}{v_0 \cos(\alpha)}$

Step 2: Vertical Motion

The vertical motion equation can be written as:

$y = y_0 + v_0 \sin(\alpha) \cdot t - \frac{1}{2} g t^2$

Where:

• $y$ is the height of the basketball hoop.
• $y_0$ is the release height.
• $g$ is the gravitational acceleration.

Given that $y = h$, we can substitute and solve for $t$ using the equation for $t$ derived from the horizontal motion:

$h = y_0 + v_0 \sin(\alpha) \cdot t - \frac{1}{2} g t^2$

Substituting $t = \frac{x}{v_0 \cos(\alpha)}$:

$h = y_0 + v_0 \sin(\alpha) \cdot \frac{x}{v_0 \cos(\alpha)} - \frac{1}{2} g \left(\frac{x}{v_0 \cos(\alpha)}\right)^2$

Simplifying, we get:

$h = y_0 + x \tan(\alpha) - \frac{g x^2}{2 v_0^2 \cos^2(\alpha)}$

Step 3: Solve for $v_0$

Now, solve for $v_0$:

$h - y_0 = x \tan(\alpha) - \frac{g x^2}{2 v_0^2 \cos^2(\alpha)}$

Rearrange to isolate $v_0^2$:

$\frac{g x^2}{2 v_0^2 \cos^2(\alpha)} = x \tan(\alpha) - (h - y_0)$

$v_0^2 = \frac{g x^2}{2 \cos^2(\alpha) \left( x \tan(\alpha) - (h - y_0) \right)}$

Finally, solve for $v_0$:

$v_0 = \sqrt{\frac{g x^2}{2 \cos^2(\alpha) \left( x \tan(\alpha) - (h - y_0) \right)}}$

Step 4: Substitute the Given Values

Given:

• $h = 3.05 \, \text{m}$
• $y_0 = 2.0 \, \text{m}$
• $x = 6.6 \, \text{m}$
• $\alpha = 50^\circ$
• $g = 10 \, \text{m/s}^2$

Substitute these values:

$v_0 = \sqrt{\frac{10 \times (6.6)^2}{2 \times \cos^2(50^\circ) \times \left( 6.6 \times \tan(50^\circ) - (3.05 - 2.0) \right)}}$

Now, let's calculate the value.The initial speed $v_0$ of the basketball when it leaves Noa's hand should be approximately 8.79 m/s for the ball to go through the hoop.

Would you like any further details or explanations on this problem?

Here are five related questions you might find interesting:

1. How does the angle of release affect the required initial speed?
2. What would happen to the initial speed if the horizontal distance to the hoop increased?
3. How would air resistance alter the required initial speed in this scenario?
4. Can this calculation be adjusted for different gravitational conditions, like on the Moon?
5. How would the initial speed change if the release height was higher?

Tip: Remember that increasing the angle of release will increase the time the ball stays in the air, but it also requires a higher initial speed to cover the same horizontal distance.