Problem 64 - College Physics

Question

A batter hits a baseball at a speed of 35.0 \(\mathrm{m} / \mathrm{s}\) and an angle of \(65.0^{\circ}\) above the horizontal. At the same instant, an outfielder 70.0 \(\mathrm{m}\) away begins running away from the batter in the line of the ball's flight, hoping to catch it. How fast must the out fielder run to catch the ball? ( (ignore air resistance, and assume the fielder catches the ball at the same height at which it left the bat.)

Step-by-Step Solution

Answer

The outfielder must run at approximately 4.66 m/s.

Step 1: Identify and Split Initial Velocity Components

First, we need to determine the horizontal and vertical components of the baseball's initial velocity. The initial speed is given as 35.0 m/s, and it is hit at an angle of 65 degrees above the horizontal. The horizontal component of the velocity, \(v_{x}\), is given by \(v_{x} = v \cdot \cos(65^{\circ})\). The vertical component, \(v_{y}\), is given by \(v_{y} = v \cdot \sin(65^{\circ})\). Where \(v = 35.0 \, \text{m/s}\).

Step 2: Calculate the Time of Flight

To find out how long the ball is in the air, we analyze the vertical motion. Using the equation for vertical displacement, which is zero since the ball lands at the same height it was hit: \(0 = v_{y} \cdot t + \frac{1}{2}(-g)t^2\). This simplifies to \(t = \frac{2v_{y}}{g}\) where \(g = 9.8\, \text{m/s}^2\). Calculate \(v_{y}\) using the expression from Step 1, then solve for \(t\).

Step 3: Determine Horizontal Distance Covered by the Ball

We calculate the horizontal distance covered by the ball during its flight time (\(t\)) using the initial horizontal velocity (\(v_{x}\)) and the formula \(x = v_{x} \cdot t\). Compute \(v_{x}\) from Step 1, then determine \(x\) using the time of flight from Step 2.

Step 4: Calculate the Required Speed of the Outfielder

The outfielder needs to cover a linear distance of \(x - 70\) meters in the same time \(t\) the ball is in the air (since he starts 70 meters away from the batter). The speed needed can therefore be calculated as \(v_{fielder} = \frac{x - 70}{t}\). Use the computed distance \(x\) and time \(t\) from the previous steps to find \(v_{fielder}\).

Key Concepts

Initial Velocity Components

When a baseball is hit at an angle, its velocity is divided into horizontal and vertical components. This makes it easier to analyze the motion of the ball. To find these components, we rely on some trigonometry. The angle at which the ball is hit is crucial here. In our problem, this angle is given as \(65^{\circ}\), and the initial speed is \(35.0\, \text{m/s}\).

The horizontal component of velocity \(v_x\) is calculated as:

• \(v_x = v \cdot \cos(65^{\circ})\)

The vertical component \(v_y\) is found by:

• \(v_y = v \cdot \sin(65^{\circ})\)

Where \(v\) represents the initial speed of the ball.

By splitting the initial velocity into these two components, we can independently analyze the projectile's motion in vertical and horizontal directions. This separation is fundamental to solving the motion problem efficiently.

Time of Flight

The time a projectile stays in the air, or the "time of flight," is determined by its vertical motion. In this case, since the ball is caught at the same height it was launched, the vertical displacement is zero.

To calculate this, we utilize the formula for the vertical motion:

• \(0 = v_y \cdot t + \frac{1}{2}(-g)t^2\)

Given that \(g\) is the acceleration due to gravity \(9.8\, \text{m/s}^2\), you can simplify the equation to:

• \(t = \frac{2v_y}{g}\)

This formula provides the time the ball remains airborne. Calculate \(v_y\) using the method outlined earlier, substitute it into this equation, and solve for \(t\).

Understanding the time of flight is crucial as it allows us to sync both the ball and the outfielder's motion in the same timeframe.

Horizontal Distance

Once we know how long the ball is in the air, we can find out how far it travels horizontally. This is straightforward if we know the horizontal velocity component \(v_x\) and the time the ball is in the air \(t\).

The formula for horizontal distance \(x\) covered is:

• \(x = v_x \cdot t\)

Since \(v_x\) came from our earlier calculations, and \(t\) is derived from the time of flight analysis, we can directly compute \(x\).

Knowing this distance helps in understanding how far the outfielder must run before catching the ball, taking into account his initial position relative to the batter.

Velocity Calculation

The final step links both the projectile’s and outfielder's motions, determining what speed the outfielder needs to maintain to catch the ball.

The outfielder starts \(70\, \text{m}\) away from the batter and must reach the position where the ball lands, which is the horizontal distance \(x\) previously calculated. The catch has to occur within the same time span \(t\) that the ball is in the air.

The required speed \(v_{\text{fielder}}\) is:

• \(v_{\text{fielder}} = \frac{x - 70}{t}\)

This formula finds the outfielder's required running speed by accounting for the additional distance needed to cover \((x - 70)\).

By performing this calculation, you ensure the parameters are met for the outfielder to successfully catch the ball, integrating all parts of the projectile motion into a cohesive solution.