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, ? = 42.0A????1, with the horizontal from a height of h = 2.20 m.
(a) What is the acceleration of the basketball at the highest point in its trajectory?
magnitude
direction —Select— upward, or downward
(b) At what speed must the player throw the basketball so that the ball goes through the hoop without striking the backboard?
m/s
The Correct Answer and Explanation is :
(a) What is the acceleration of the basketball at the highest point in its trajectory?
At the highest point in its trajectory, the basketball has no vertical velocity. However, the vertical acceleration due to gravity is still acting on it, pulling it downward.
- Magnitude: The magnitude of the acceleration is constant throughout the entire flight of the ball and is equal to the acceleration due to gravity. This value is approximately ( 9.81 \, \text{m/s}^2 ).
- Direction: The direction of the acceleration is downward since gravity pulls the ball toward the ground.
Answer for (a):
- Magnitude: ( 9.81 \, \text{m/s}^2 )
- Direction: Downward
(b) At what speed must the player throw the basketball so that the ball goes through the hoop without striking the backboard?
To solve this, we will use projectile motion equations. The ball’s initial velocity can be split into two components:
- The horizontal component (( v_{0x} )) and
- The vertical component (( v_{0y} )).
Given:
- Horizontal distance (range): ( x = 10.0 \, \text{m} )
- Initial vertical height: ( y_0 = 2.20 \, \text{m} )
- Height of the basket: ( y_{\text{basket}} = 3.05 \, \text{m} )
- Launch angle: ( \theta = 42^\circ )
- Gravity: ( g = 9.81 \, \text{m/s}^2 )
We know that the horizontal velocity is constant throughout the flight because there is no horizontal acceleration. The equations for projectile motion are:
- Horizontal Motion:
[
x = v_0 \cdot \cos(\theta) \cdot t
] - Vertical Motion:
[
y = y_0 + v_0 \cdot \sin(\theta) \cdot t – \frac{1}{2} g t^2
]
To find the velocity ( v_0 ), we need to eliminate time (( t )) from the equations. First, solve the horizontal motion equation for time (( t )):
[
t = \frac{x}{v_0 \cos(\theta)}
]
Substitute this expression for ( t ) into the vertical motion equation:
[
y_{\text{basket}} = y_0 + v_0 \cdot \sin(\theta) \cdot \frac{x}{v_0 \cdot \cos(\theta)} – \frac{1}{2} g \left( \frac{x}{v_0 \cdot \cos(\theta)} \right)^2
]
Simplify this equation and solve for ( v_0 ). Using this approach, we can find the required initial velocity to make the basketball go through the hoop.
Let’s calculate this:
[
y_{\text{basket}} = y_0 + x \cdot \tan(\theta) – \frac{g x^2}{2 v_0^2 \cdot \cos^2(\theta)}
]
Substitute the known values and solve for ( v_0 ):
[
3.05 = 2.20 + 10.0 \cdot \tan(42^\circ) – \frac{9.81 \cdot 10.0^2}{2 v_0^2 \cdot \cos^2(42^\circ)}
]
I’ll calculate this now.
The required initial velocity for the basketball to go through the hoop without striking the backboard is approximately ( 10.44 \, \text{m/s} ).
Explanation:
We derived the equation based on the projectile motion principles and then solved for the initial velocity ( v_0 ) that would allow the ball to travel a horizontal distance of 10.0 meters and reach the height of 3.05 meters at the basket.
The negative value for velocity is discarded, as velocity cannot be negative in this context. Therefore, the player must throw the ball with a speed of around ( 10.44 \, \text{m/s} ) at an angle of ( 42^\circ ) to ensure it reaches the basket without hitting the backboard.