Problem 136 - Fundamentals of Physics

Question

A baseball is hit at Fenway Park in Boston at a point \(0.762 \mathrm{~m}\) above home plate with an initial velocity of \(33.53 \mathrm{~m} / \mathrm{s}\) directed \(55.0^{\circ}\) above the horizontal. The ball is observed to clear the \(11.28\) -m-high wall in left field (known as the "green monster") \(5.00 \mathrm{~s}\) after it is hit, at a point just inside the left-field foulline pole. Find (a) the horizontal distance down the left-field foul line from home plate to the wall; (b) the vertical distance by which the ball clears the wall; (c) the horizontal and vertical displacements of the ball with respect to home plate \(0.500 \mathrm{~s}\) before it clears the wall.

Step-by-Step Solution

Answer

(a) 96.10 m; (b) 54.48 m; (c) 86.49 m, 55.47 m

Step 1: Break Down the Initial Velocity

The initial velocity is given as 33.53 m/s at an angle of 55.0° above the horizontal. We will split this into horizontal and vertical components using trigonometry.The horizontal component of the initial velocity is obtained using cosine:\[ v_{x0} = v_0 \cos(\theta) = 33.53 \cos(55.0^{\circ}) = 19.22 \mathrm{~m/s} \]The vertical component of the initial velocity is obtained using sine:\[ v_{y0} = v_0 \sin(\theta) = 33.53 \sin(55.0^{\circ}) = 27.50 \mathrm{~m/s} \]

Step 2: Calculate Horizontal Distance to the Wall

The horizontal distance traveled by the baseball is given by the product of the horizontal velocity and time. The time given is 5.00 seconds.\[ x = v_{x0} \times t = 19.22 \times 5.00 = 96.10 \mathrm{~m} \]

Step 3: Calculate the Height of the Ball at 5 Seconds

We need to find the vertical position of the ball at 5 seconds. The motion in the vertical direction is subject to gravitational acceleration.Use the kinematic equation:\[ y = y_0 + v_{y0}t - \frac{1}{2}gt^2 \]Substitute the known values (\(y_0 = 0.762 \mathrm{~m}\),\(v_{y0} = 27.50 \mathrm{~m/s}\),\(g = 9.81 \mathrm{~m/s^2}\), and\( t = 5.00 \mathrm{~s} \)):\[ y = 0.762 + 27.50 \times 5.00 - 0.5 \times 9.81 \times 5.00^2 = 65.76 \mathrm{~m} \]

Step 4: Determine Vertical Clearance Over the Wall

Now, compute the clearance by subtracting the wall height from the calculated height of the ball after 5 seconds:\[ \text{Clearance} = 65.76 - 11.28 = 54.48 \mathrm{~m} \]

Step 5: Find Horizontal and Vertical Displacement at 4.5 Seconds

To find where the ball is 0.5 seconds before reaching the wall (at 4.5 seconds), compute the horizontal and vertical displacements at that time.For horizontal displacement:\[ x = v_{x0} \times t = 19.22 \times 4.5 = 86.49 \mathrm{~m} \]For vertical displacement:\[ y = y_0 + v_{y0} \times t - 0.5 \times g \times t^2 \]Substitute known values for 4.5 seconds:\[ y = 0.762 + 27.50 \times 4.5 - 0.5 \times 9.81 \times 4.5^2 = 55.47 \mathrm{~m} \]

Key Concepts

Kinematics

Kinematics is a branch of physics that studies the motion of objects without considering the forces that cause the motion. This field focuses on understanding concepts like displacement, velocity, and acceleration. In projectile motion, kinematics is particularly useful as it allows us to track the path of a projectile as it travels through space.

By applying kinematic equations, we can determine the ballistic trajectory of an object like a baseball. These equations help us predict features such as the horizontal distance it will travel and its height at a specific time. In our problem, we used kinematic equations to determine how the components of the initial velocity influence the baseball's flight path, which is a crucial aspect of projectile motion.

Initial velocity components

Understanding the initial velocity components is essential in analyzing projectile motion. Initially, the velocity is given in a specific direction. We need to break it down into its horizontal and vertical components using trigonometry.

The horizontal component, calculated using the cosine of the launch angle, determines how fast the projectile will cover distance along the ground. It remains constant throughout the flight since there are no horizontal forces acting on the projectile:

• Horizontal velocity: \( v_{x0} = v_0 \cos(\theta) = 19.22 \mathrm{~m/s} \)

The vertical component, derived using the sine of the launch angle, dictates how high and for how long the projectile will rise against gravity, which causes it to decelerate upwards and accelerate downwards:

• Vertical velocity: \( v_{y0} = v_0 \sin(\theta) = 27.50 \mathrm{~m/s} \)

These components are critical for predicting the projectile's trajectory and motion behavior over time.

Horizontal displacement

Horizontal displacement measures how far an object travels horizontally over time. It is determined by the initial horizontal velocity and the time of travel. Since there is no horizontal acceleration in ideal projectile motion, the horizontal velocity remains constant, and thus the displacement is simple to calculate:

• Horizontal distance: \( x = v_{x0} \cdot t \)

In our example, the baseball's horizontal distance was computed for two different times:

• Initial time (5 s): \( x = 19.22 \times 5.00 = 96.10 \mathrm{~m} \)
• Prior to reaching the wall (4.5 s): \( x = 19.22 \times 4.5 = 86.49 \mathrm{~m} \)

Understanding horizontal displacement is key for analyzing how far a projectile will land, a question that is quite frequent in sports and engineering problems alike.

Vertical clearance

Vertical clearance is concerned with how much higher or lower a projectile is compared to a reference point, such as a wall or building. This involves calculating the vertical position at a particular time and comparing it to the height of an obstacle.

For the baseball, we used this to determine whether it clears Fenway Park's wall by finding the ball's height at 5 seconds:

• Vertical position: \( y = y_0 + v_{y0} \cdot t - 0.5 \cdot g \cdot t^2 \)

Plugging in the values, the height was calculated to be 65.76 m. By subtracting the wall's height (11.28 m), we find the clearance:

• Clearance: \( 65.76 - 11.28 = 54.48 \mathrm{~m} \)

This calculation is vital in many practical situations, such as ensuring a launched object clears an obstacle safely.