Problem 31 - Fundamentals of Physics

Question

In a jump spike, a volleyball player slams the ball from overhead and toward the opposite floor. Controlling the angle of the spike is difficult. Suppose a ball is spiked from a height of \(2.30\) \(\mathrm{m}\) with an initial speed of \(20.0 \mathrm{~m} / \mathrm{s}\) at a downward angle of \(18.00^{\circ} .\) How much farther on the opposite floor would it have landed if the downward angle were, instead, \(8.00^{\circ} ?\)

Step-by-Step Solution

Answer

The volleyball lands farther with the \(8.00^{\circ}\) angle because it travels more horizontally before landing.

Step 1: Understand the Problem

We need to calculate how much further a volleyball lands when the spiking angle is decreased from \(18.00^{\circ}\) to \(8.00^{\circ}\). We will use projectile motion equations for this. The ball is projected from a height of \(2.30 \text{ m}\) with an initial speed of \(20.0 \text{ m/s}\).

Step 2: Break Down Initial Velocity into Components

For both angles, break down the initial velocity into horizontal and vertical components using \( v_{0x} = v_0 \cdot \cos(\theta) \) and \( v_{0y} = v_0 \cdot \sin(\theta) \), where \( \theta \) is the angle with the horizontal. Calculate \( v_{0x} \) and \( v_{0y} \) for both \(18.00^{\circ}\) and \(8.00^{\circ}\) angles.

Step 3: Calculate Time of Flight for Each Angle

Use the equation for vertical motion \( y = v_{0y} \cdot t + \frac{1}{2} g \cdot t^2 \) to calculate the time \( t \) it takes for the ball to hit the floor for both angles. Set \( y = -2.30 \text{ m} \) (since it's 2.30 m below the release point) and solve the quadratic equation for \( t \).

Step 4: Calculate Horizontal Range for Each Angle

Once \( t \) is found for each angle, calculate the horizontal range using \( x = v_{0x} \cdot t \). This gives you the distance from the point of projection to where the ball lands.

Step 5: Compare the Two Ranges

Subtract the horizontal range for angle \(18.00^{\circ}\) from that for angle \(8.00^{\circ}\) to find out how much farther the ball lands with the smaller angle.

Key Concepts

Initial Velocity Components

In projectile motion, understanding the initial velocity components is crucial. The initial velocity of any projectile is typically given as a value combined with an angle. This is the speed and the direction at which the projectile is launched. To analyze this, we need to break it down into two components:

• Horizontal Component: This is found using the formula \( v_{0x} = v_0 \cdot \cos(\theta) \). It determines how fast the projectile is moving across the ground.
• Vertical Component: This is found using the formula \( v_{0y} = v_0 \cdot \sin(\theta) \). This determines how fast the projectile is moving up and or down.

For instance, with the volleyball, if it's hit with an initial velocity of \( 20 \ \text{m/s} \) at an angle of \(18^{\circ}\), the horizontal and vertical components will differ compared to if the angle were \(8^{\circ}\). Calculating these components accurately helps in understanding the behavior of the volleyball's motion.

Time of Flight

The time of flight refers to the total time a projectile remains in the air. For vertically moving objects, we use the vertical motion equation \( y = v_{0y} \cdot t + \frac{1}{2} g \cdot t^2 \). Here, \( y \) is the vertical displacement, which in our scenario is \(-2.30 \, \text{m}\) because the volleyball starts at \(2.30 \, \text{m}\) and ends at ground level.To determine the time of flight, we solve this equation for \( t \), treating it as a quadratic equation. Remember, \( g \) is the acceleration due to gravity (approximately \( -9.81 \, \text{m/s}^2 \)). Calculating time of flight helps us understand how long it will take before the projectile reaches the ground from its release point.

Horizontal Range

Once you know the time of flight, finding the horizontal range is straightforward. The horizontal range is the total horizontal distance the projectile covers during its flight. To find this, use the equation:\[ x = v_{0x} \cdot t\]where \( x \) is the horizontal distance, \( v_{0x} \) is the horizontal component of the initial velocity, and \( t \) is the time of flight. This distance changes based on the angle of launch, like in our volleyball problem where different angles (\(18^{\circ}\) and \(8^{\circ}\)) result in different ranges. Calculating the horizontal range helps us predict where the projectile will land.

Vertical Motion Equations

Vertical motion equations are essential for analyzing the projectile's motion in the vertical direction. The primary equation used in this context is:\[y = v_{0y} \cdot t + \frac{1}{2} g \cdot t^2\]This equation gives us the vertical displacement \( y \) at any time \( t \) during the flight. The term \( v_{0y} \cdot t \) represents the influence of the initial vertical velocity, while \( \frac{1}{2} g \cdot t^2 \) accounts for the acceleration due to gravity.

• Negative values indicate motion below the original position.
• Positive values indicate motion above the initial position (not common if landing is below release point).

Understanding these equations allows us to calculate times, velocities, and positions of projectiles at various points in their flight. In our problem, it helps determine how deep the volleyball will land below its initial hitting point.