Question

A baseball leaves the hand of a pitcher 6 vertical feet above home plate and \(60 \mathrm{ft}\) from home plate. Assume that the coordinate axes are oriented as shown in the figure. a. In the absence of all forces except gravity, assume that a pitch is thrown with an initial velocity of \(\langle 130,0,-3\rangle \mathrm{ft} / \mathrm{s}\) (about \(90 \mathrm{mi} / \mathrm{hr}\) ). How far above the ground is the ball when it crosses home plate and how long does it take for the pitch to arrive? b. What vertical velocity component should the pitcher use so that the pitch crosses home plate exactly \(3 \mathrm{ft}\) above the ground? c. A simple model to describe the curve of a baseball assumes that the spin of the ball produces a constant sideways acceleration (in the \(y\) -direction) of \(c \mathrm{ft} / \mathrm{s}^{2}\). Assume a pitcher throws a curve ball with \(c=8 \mathrm{ft} / \mathrm{s}^{2}\) (one-fourth the acceleration of gravity). How far does the ball move in the \(y\) -direction by the time it reaches home plate, assuming an initial velocity of (130,0,-3) ft/s? d. In part (c), does the ball curve more in the first half of its trip to the plate or in the second half? How does this fact affect the batter? e. Suppose the pitcher releases the ball from an initial position of (0,-3,6) with initial velocity \((130,0,-3) .\) What value of the spin parameter \(c\) is needed to put the ball over home plate passing through the point (60,0,3)\(?\)

Short answer

Answer: The required initial vertical velocity component is approximately -1.62 ft/s.

Step-by-step solution

Find the position vector function

To work with this problem, we first need to find the position vector function. We have the initial velocity vector \(\langle 130,0,-3\rangle \mathrm{ft} / \mathrm{s}\) and we know that the acceleration due to gravity is downwards with a value of \(32\mathrm{ft} / \mathrm{s}^{2}\). So our acceleration vector is \(\langle 0,0,-32\rangle \mathrm{ft} / \mathrm{s}^{2}\). Using this information, we can write the position vector function for the ball: \(r(t) = \langle 0,0,6\rangle + \langle 130,0,-3\rangle t + \frac{1}{2}\langle 0,0,-32\rangle t^2\)

Evaluate the x-component position

Since the initial position is \(6\mathrm{ft}\), the horizontal distance between the pitcher and home plate is \(60\mathrm{ft}\). We need to find the time it takes for the ball to travel \(60\mathrm{ft}\). To do this, we set the x-component of the position vector equal to 60: \(60 = 130t\) Now solve for \(t\): \(t = \frac{60}{130} = \frac{3}{6.5}\)

Find the ball's z-coordinate at the home plate

Now that we have the time it takes for the pitch to arrive (\(t=\frac{3}{6.5}\)), we can plug this value back into our position vector function to find how far above the ground the baseball is: \(z(t) = 6 - 3\left(\frac{3}{6.5}\right) - 16\left(\frac{3}{6.5}\right)^2\)

Simplify the expression to get the required height z(t)

Solve for the resulting height of the baseball: \(z(t) = 6 - \frac{9}{6.5} - \frac{144}{6.5^2}\) \(z(t) \approx 3.27 \mathrm{ft}\) So, the ball is approximately \(3.27\mathrm{ft}\) above the ground when it crosses the home plate and it takes \(\frac{3}{6.5} \mathrm{s}\) for the pitch to arrive. #b. Finding the required initial vertical velocity component#

Set up the equation to find the required initial vertical velocity component

Let's denote the required initial vertical velocity component as \(v_{z}\). The position vector function will now be: \(z(t) = 6 + v_{z}t - 16t^2\) Since we want the ball to cross the home plate exactly \(3\mathrm{ft}\) above the ground, we set \(z(t)\) to 3: \(3 = 6 + v_{z}\left(\frac{3}{6.5}\right) - 16\left(\frac{3}{6.5}\right)^2\)

Solve for the initial vertical velocity component \(v_{z}\)

Rearrange the equation to find the initial vertical velocity component \(v_{z}\): \(v_{z} = \left(3 - 6 + 16\left(\frac{9}{6.5^2}\right)\right)\frac{6.5}{3}\) \(v_{z} \approx -1.62\) ft/s Thus, the pitcher should use a vertical velocity component of approximately \(-1.62\mathrm{ft/s}\) to ensure the pitch crosses the home plate \(3\mathrm{ft}\) above the ground. #c. Finding the ball's movement in the y-direction#

Find the y-coordinate position function r_y(t)

In the presence of constant sideways acceleration (in the y-direction) of 8ft/s², we can write the position function for the ball's y-component as: \(r_y(t) = 0 + 0t + \frac{1}{2}(8)t^2\)

Compute the ball's y-coordinate at the home plate

Now, evaluate the y-coordinate of the ball when it reaches the home plate by substituting the time \(t = \frac{3}{6.5}\): \(r_y \left(\frac{3}{6.5}\right) = \frac{1}{2}(8)\left(\frac{3}{6.5}\right)^2\) \(r_y \left(\frac{3}{6.5}\right) \approx 1.94\mathrm{ft}\) Therefore, the ball moves approximately \(1.94\mathrm{ft}\) in the y-direction by the time it reaches home plate. #d. Whether the ball curves more in the first half or second half#

Compute the y-coordinate midway t = \frac{3}{13}

Since the time taken to reach the home plate is \(\frac{3}{6.5}\), we can find the ball's y-coordinate midway (at t = \(\frac{3}{13}\)) using the y-component position function: \(r_y \left(\frac{3}{13}\right) = \frac{1}{2}(8)\left(\frac{3}{13}\right)^2\) \(r_y \left(\frac{3}{13}\right) \approx 0.47\mathrm{ft}\)

Compare the y-coordinate midway and y-coordinate at home plate

Upon comparing, we see that the ball curves more in the second half of its trip as \(1.94 - 0.47 = 1.47\) ft. In the second half, the ball moves approximately \(1.47\mathrm{ft}\) in the y-direction, while it moves \(0.47\mathrm{ft}\) in the first half. This effect makes it more challenging for the batter, as the batter has less time and information to anticipate where the ball will be when it arrives at the home plate. #e. Finding the value of spin parameter c to pass through (60,0,3)#

Compute the initial y-position (0,-3,6)

We are given the initial position of the ball as \((0, -3, 6)\). So, we can write the position vector function for the ball as follows: \(r(t) = \langle 0,-3,6\rangle + \langle 130,0,-3\rangle t + \frac{1}{2}\langle 0,c,-32\rangle t^2\)

Plug in the desired position (60,0,3) and t = \frac{3}{6.5}

To find the value of spin parameter \(c\), we plug in the desired position \((60, 0, 3)\) and the time \(t = \frac{3}{6.5}\): \(\langle 60,0,3\rangle = \langle 0,-3,6\rangle + \left \langle 130,0,-3\right\rangle \left(\frac{3}{6.5} \right) + \frac{1}{2}\left\langle 0,c,-32\right\rangle \left(\frac{3}{6.5} \right)^2\)

Solve for the spin parameter \(c\)

Now, solve for the spin parameter \(c\): \(c = \frac{6\left(\frac{3}{6.5}\right)^2}{\left(\frac{3}{6.5}\right)^2}=6\) So, the spin parameter \(c\) required for the ball to pass through \((60,0,3)\) is \(6 \mathrm{ft/s^2}\).

Key concepts

Initial Velocity

Initial velocity is a key parameter in projectile motion, defining the speed and direction an object has when it begins its path. In this problem, the baseball is thrown with an initial velocity vector of \( \langle 130,0,-3\rangle \) ft/s. This vector indicates that the ball travels:

• Horizontally at 130 ft/s, signifying its speed towards the home plate.
• With no initial movement in the y-direction, which means the ball starts its journey straight without any sideways curve.
• Vertically downwards at -3 ft/s, showing it is moving slightly downward right from the start.

The initial velocity impacts how far and how high the ball travels. Adjusting any component of this vector alters the ball's trajectory significantly.

Gravity

Gravity is the force pulling objects towards the earth. For objects in free fall like the baseball, gravity directly influences their motion. In this problem, gravity is modeled as an acceleration vector \( \langle 0,0,-32 \rangle \) ft/s², which only affects the vertical component of the ball's motion.This gravitational pull causes:

• The ball to gain downward velocity as it travels.
• Changes in the ball’s height over time, making the trajectory a curved path.

In calculations, this constant acceleration is crucial in determining both the time it takes the ball to hit a certain point and the position it will be in when it gets there. To predict the path, gravity needs to be factored in as it diminishes the vertical velocity component over time.

Acceleration

Acceleration in projectile motion refers to the rate of change of velocity. While gravity primarily affects the vertical acceleration, additional forces like spin can affect the motion. Here, a sideways acceleration component is considered to model how spin affects the ball’s flight, particularly in parts (c) and (e) of the problem where a sideways acceleration \( c = 8 \) ft/s² is introduced.This sideways acceleration component impacts the ball’s path by:

• Causing it to drift sideways as it travels horizontally.
• Making a simple straight pitch curve towards one side, influencing a batter's perception.

Such effects can be particularly prominent in sports like baseball where the art of pitching heavily relies on manipulating acceleration through spin to deceive batters.

Position Vector Function

The position vector function describes the ball's trajectory at any given point in time. Combining initial velocity and acceleration, it gives a dynamic representation of motion. For the baseball, the position vector function is:\[ r(t) = \langle 0,0,6 \rangle + \langle 130,0,-3 \rangle t + \frac{1}{2} \langle 0,0,-32 \rangle t^2 \]This formula helps determine:

• How the x, y, and z positions of the ball change over time.
• At what exact distance and height the ball will be after any elapsed time \( t \).

To solve parts of this problem, such as when the ball reaches the plate or how high it is above ground at that point, substituting appropriate \( t \) values into this function allows us to track its precise movement.