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, è = 42.8°, 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 and its direction?

To solve this problem, we need to analyze the projectile motion of the basketball. Let's break it down step by step.

Step 1: Given data:
- Horizontal distance from the player to the basket (range): 10.0 m
- Height of the basket (H): 3.05 m
- Angle of the shot (θ): 42.8°
- Initial height of the ball (h): 2.04 m

Step 2: Calculate the initial velocity components:
The initial velocity of the basketball can be divided into horizontal and vertical components.
- Horizontal component (Vx): Vx = V * cos(θ)
- Vertical component (Vy): Vy = V * sin(θ)

Step 3: Calculate the time it takes for the basketball to reach its highest point:
The time it takes for the ball to reach its highest point can be determined by finding the time it takes for the vertical component of the velocity (Vy) to reach zero.
- Using the kinematic equation: Vy = V0y + at
- At the highest point, Vy = 0 m/s, so a = -g (acceleration due to gravity) and V0y = Vy (initial vertical velocity)
- Rearranging the equation: 0 = Vy - gt
- Solving for t: t = Vy / g

Step 4: Calculate the acceleration at the highest point:
The acceleration of the basketball at the highest point is equal to the acceleration due to gravity (g), but with the opposite sign (-g). This is because gravity is constantly acting downward and decelerating the ball as it moves upward.
- So, the acceleration at the highest point = -g

Therefore, the acceleration of the basketball at the highest point in its trajectory is equal to the acceleration due to gravity (-9.8 m/s²) and its direction is downward.