A basketball player is standing on the floor 10.0 m from the basket as in the figure below. The height of the basket is H = 3.05 m, and he shoots the ball at an angle, 𝜃 = 37.6°, with the horizontal from a height of h = 2.04 m.

(a) What is the acceleration of the basketball at the highest point in its trajectory?
magnitude	  m/s2
direction	

(b) At what speed must the player throw the basketball so that the ball goes through the hoop without striking the backboard?
  m/s

To solve the problem, we can break it into two parts: 
1. The acceleration of the basketball at the highest point in its trajectory.
2. The initial speed required for the ball to go through the hoop without hitting the backboard.

### Part (a): The Acceleration of the Basketball at the Highest Point

The basketball is in projectile motion, and the only force acting on it after it is shot is gravity. This means that the **acceleration of the ball at the highest point of its trajectory** is simply the acceleration due to gravity.

- The magnitude of the acceleration due to gravity is always \( 9.81 \, \text{m/s}^2 \).
- The **direction** of the acceleration is vertically downward.

At the highest point, the vertical velocity of the ball is zero, but the horizontal velocity remains constant. However, the **acceleration remains downward** because the only force acting on the basketball is gravity.

Thus, the answers are:

- Magnitude of acceleration: \( 9.81 \, \text{m/s}^2 \)
- Direction: Downward

### Part (b): Speed Required to Throw the Basketball to the Hoop

We need to determine the initial velocity \( v_0 \) that the player must throw the basketball at, so that it reaches the hoop.

Let's define the known values:
- Distance to the hoop: \( x = 10.0 \, \text{m} \)
- Height of the hoop: \( H = 3.05 \, \text{m} \)
- Release height: \( h = 2.04 \, \text{m} \)
- Launch angle: \( \theta = 37.6^\circ \)
- Gravitational acceleration: \( g = 9.81 \, \text{m/s}^2 \)

We will apply the **kinematic equations** for projectile motion. We can split the motion into horizontal and vertical components.

#### 1. Horizontal motion:
The horizontal velocity is given by:
\[
v_{x0} = v_0 \cos \theta
\]
The horizontal distance traveled is:
\[
x = v_{x0} t = v_0 \cos \theta \cdot t
\]
Thus, the time \( t \) it takes for the ball to travel the distance of 10 meters is:
\[
t = \frac{x}{v_0 \cos \theta}
\]

#### 2. Vertical motion:
The vertical velocity is given by:
\[
v_{y0} = v_0 \sin \theta
\]
The vertical displacement \( y \) from the release point to the hoop is:
\[
y = H - h = 3.05 \, \text{m} - 2.04 \, \text{m} = 1.01 \, \text{m}
\]
We apply the vertical displacement equation:
\[
y = v_{y0} t - \frac{1}{2} g t^2
\]
Substitute \( v_{y0} = v_0 \sin \theta \) and the expression for \( t \):
\[
1.01 = (v_0 \sin \theta) t - \frac{1}{2} g t^2
\]

At this point, we have two equations:

1. \( t = \frac{x}{v_0 \cos \theta} \)
2. \( 1.01 = (v_0 \sin \theta) t - \frac{1}{2} g t^2 \)

We can substitute the expression for \( t \) from the first equation into the second equation and solve for \( v_0 \). Let's calculate that.It seems like I can’t do more advanced data analysis right now. Please try again later. If you'd like, I can guide you step-by-step on how to solve this manually or answer any other related questions.

Would you like me to walk you through the solution process? Here are five questions that build on this problem:

1. What is the total time the ball is in the air during the shot?
2. How high does the basketball travel at its highest point in this shot?
3. What would the player's required speed be if the angle was increased to 45°?
4. How does changing the release height affect the required speed?
5. What would happen to the projectile motion if air resistance was considered?

**Tip:** For projectile motion problems, always separate the horizontal and vertical motions, as they are independent of each other!

A basketball player is standing on the floor 10.0 m from the basket. The height of the basket is H = 3.05 m, and he shoots the ball at an angle of θ = 37.6° from a height of h = 2.04 m. (a) What is the acceleration of the basketball at the highest point in its trajectory? (b) At what speed must the player throw the basketball so that the ball goes through the hoop without striking the backboard?

Learn how to calculate the acceleration of a basketball at the highest point of its trajectory and determine the initial speed required for a perfect basketball shot. This problem explores concepts of projectile motion, using kinematic equations and trigonometry, ideal for Grades 10-12 physics students.

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