Problem 67 - College Physics

Question

A 2.05 -m-tall basketball player takes a shot when he is \(6.02 \mathrm{~m}\) from the basket (at the three-point line). If the launch angle is \(25^{\circ}\) and the ball was launched at the level of the player's head, what must be the release speed of the ball for the player to make the shot? The basket is \(3.05 \mathrm{~m}\) above the floor.

Step-by-Step Solution

Answer

The release speed must be approximately 8.15 m/s.

Step 1: Identify the Known Values

The problem provides several key pieces of information:- Initial height of the ball, which is the height of the player: \( h_0 = 2.05\,\mathrm{m} \)- Distance from player to basket: \( d = 6.02\,\mathrm{m} \)- Launch angle: \( \theta = 25^{\circ} \)- Height of the basket: \( h = 3.05\, \mathrm{m} \)

Step 2: Determine Vertical and Horizontal Components

To solve for the release speed, we'll separate the launch velocity into horizontal and vertical components:- Horizontal component: \( v_{0x} = v_0 \cos \theta \)- Vertical component: \( v_{0y} = v_0 \sin \theta \)

Step 3: Apply Projectile Motion Formula for Horizontal Motion

We need the time it takes for the ball to travel horizontally to the basket:\[ t = \frac{d}{v_{0x}} = \frac{6.02}{v_0 \cos 25^{\circ}} \]

Step 4: Apply Projectile Motion Formula for Vertical Motion

Using the vertical motion formula: \[ h = h_0 + v_{0y}t - \frac{1}{2}gt^2 \]Substitute the known values:\[ 3.05 = 2.05 + v_0 \sin 25^{\circ} \left( \frac{6.02}{v_0 \cos 25^{\circ}} \right) - \frac{1}{2}(9.81)\left(\frac{6.02}{v_0 \cos 25^{\circ}}\right)^2 \]

Step 5: Solve the Equation for Release Speed

Re-arrange the equation to find \( v_0 \):\[ 3.05 - 2.05 = v_0 \sin 25^{\circ} \left( \frac{6.02}{v_0 \cos 25^{\circ}} \right) - \frac{1}{2}(9.81)\left(\frac{6.02}{v_0 \cos 25^{\circ}}\right)^2 \]Solve for \( v_0 \) using algebraic methods or a numerical approach.

Key Concepts

Launch Angle

The launch angle is the direction angle at which an object is projected into the air. In the context of projectile motion, such as a basketball shot, the angle of launch significantly influences the motion path of the projectile. The launch angle is crucial because it affects both the horizontal and vertical distances the object will travel.
When an object is launched at a low angle, such as 25 degrees in our example, it travels more horizontally. This can increase the speed needed to cover the same horizontal distance compared to a higher angle, where the object might travel a longer vertical distance and follow a parabolic path.
Understanding the effects of different launch angles is crucial in determining the right speed and angle to achieve a successful shot or throw. Keep in mind:

• Launch angles of 0 or near 0 result in maximum horizontal speed but minimal vertical movement.
• A 45-degree angle is often ideal for maximizing range in a vacuum, but varying the angle depending on the situation can also optimize performance depending on limiting factors such as height or air resistance.
• Angles above 45 degrees usually focus on reaching greater heights rather than maximizing horizontal distance.

Vertical and Horizontal Components

Projectile motion can be understood by splitting it into two independent components: vertical and horizontal. This decomposition is fundamental when analyzing or solving projectile motion problems, including calculating the appropriate release speed.
To distinguish these components, we use trigonometric functions based on the launch angle:

• Horizontal Component: Represented by \( v_{0x} = v_0 \cos(\theta) \), this determines the speed at which the projectile moves across the ground. It remains constant during the flight, assuming air resistance is negligible.
• Vertical Component: Expressed as \( v_{0y} = v_0 \sin(\theta) \), this dictates how high or low the object moves during its flight. Gravity impacts this component, making it decelerate as the object goes up and accelerate as it comes down.

Understanding these components makes it easier to analyze the path of the projectile. For instance, knowing the vertical component of velocity helps predict how high the projectile will reach, whereas the horizontal component guides us on how far it will travel.

Release Speed

Release speed is the initial speed a projectile needs to be thrown to achieve a desired trajectory. In this exercise, we need to calculate the speed required for a ball to cover a specific distance and reach a certain height.
To determine the release speed, we calculate it using the properties of projectile motion equations. In our example, we need the basketball to travel 6.02 meters horizontally while ascending from the player's height of 2.05 meters to hit a basket at 3.05 meters above the floor.
The kinematic equations allow us to integrate both the horizontal and vertical components along with gravitational influence to find this speed. In practical terms, we solve it as follows:

• Calculate the time the ball travels horizontally based on the distance and horizontal speed.
• Use this time to analyze how high the ball will rise using vertical speed and gravitational influence.
• Combine these calculations to determine the speed at which the ball should be released to ensure it travels the full distance and all conditions for successful contact with the target are met.

When looking for a release speed, always remember it's a combination of phenomena acting together: initial speed, launch angle, distance, and the effect of gravity should be carefully orchestrated for the desired outcome.