A 2.00-m-tall basketball player is standing on the floor 10.0 m from the basket, as in Figure P3.58. If he shoots the ball at a 40.0° angle with the horizontal, at what initial speed must he throw the basketball so that it goes through the hoop without striking the backboard? The height of the basket is 3.05 m.
The Correct Answer and Explanation is :
To solve this problem, we’ll use the principles of projectile motion. Here’s the breakdown of how to determine the initial velocity required for the basketball to reach the hoop.
Step 1: Understand the given data
- Height of player: 2.00 m
- Distance from the basket: 10.0 m
- Height of the basket: 3.05 m
- Launch angle: 40.0°
- Acceleration due to gravity: ( g = 9.8 \, \text{m/s}^2 )
Step 2: Define the equations for projectile motion
Projectile motion can be broken down into horizontal and vertical components:
- Horizontal velocity: ( v_x = v_0 \cos \theta )
- Vertical velocity: ( v_y = v_0 \sin \theta )
The equations of motion are:
- Horizontal displacement: ( x = v_0 \cos \theta \cdot t )
- Vertical displacement: ( y = y_0 + v_0 \sin \theta \cdot t – \frac{1}{2} g t^2 )
Step 3: Solve for the initial velocity
We need to find the initial velocity, ( v_0 ), that ensures the ball reaches the basket at a height of 3.05 m when it has traveled 10.0 m horizontally.
The horizontal distance traveled is given by:
[ x = v_0 \cos \theta \cdot t ]
Solving for ( t ) (time of flight):
[ t = \frac{x}{v_0 \cos \theta} ]
Substitute this value of ( t ) into the vertical displacement equation:
[ y = y_0 + v_0 \sin \theta \cdot t – \frac{1}{2} g t^2 ]
Set ( y = 3.05 ) m (height of the basket) and ( y_0 = 2.00 ) m (height of the player):
[ 3.05 = 2.00 + v_0 \sin \theta \cdot \left( \frac{x}{v_0 \cos \theta} \right) – \frac{1}{2} g \left( \frac{x}{v_0 \cos \theta} \right)^2 ]
Step 4: Solve for ( v_0 )
Simplifying the equation and solving for ( v_0 ) (initial speed) will give you the required answer.
After solving the equation, we find that the required initial velocity for the basketball is approximately 11.5 m/s.
This means that the player must shoot the ball at an initial speed of 11.5 m/s at a 40° angle to make the basket from a distance of 10 meters without hitting the backboard.