Question

A softball is thrown from the origin of an \(x-y\) coordinate system with an initial speed of \(18 \mathrm{~m} / \mathrm{s}\) at an angle of \(35^{\circ}\) above the horizontal. (a) Find the \(x\) and \(y\) positions of the softball at the times \(t=0.50 \mathrm{~s}, 1.0 \mathrm{~s}, 1.5 \mathrm{~s}\), and \(2.0 \mathrm{~s}\). (b) Plot the results from part (a) on an \(x-y\) coordinate system, and sketch the parabolic curve that passes through them.

Short answer

Positions are calculated; plot them to reveal the parabolic path.

Step-by-step solution

Determine Initial Velocity Components

The initial velocity of the softball can be broken down into two components: horizontal and vertical. The horizontal component \( v_{x,0} \) is given by \( v_{x,0} = v_0 \cdot \cos(\theta) \), while the vertical component \( v_{y,0} \) is given by \( v_{y,0} = v_0 \cdot \sin(\theta) \). Given: \( v_0 = 18 \mathrm{~m/s} \) and \( \theta = 35^{\circ} \). Calculate the components:\[ v_{x,0} = 18 \cdot \cos(35^{\circ}) \approx 14.74 \mathrm{~m/s} \]\[ v_{y,0} = 18 \cdot \sin(35^{\circ}) \approx 10.32 \mathrm{~m/s} \]

Calculate the X Position at Different Times

The horizontal position \( x(t) \) is determined using the equation \( x(t) = v_{x,0} \cdot t \). Calculate the x-position for each given time:- At \( t = 0.5 \mathrm{~s} \): \[ x(0.5) = 14.74 \cdot 0.5 = 7.37 \mathrm{~m} \]- At \( t = 1.0 \mathrm{~s} \): \[ x(1.0) = 14.74 \cdot 1.0 = 14.74 \mathrm{~m} \]- At \( t = 1.5 \mathrm{~s} \): \[ x(1.5) = 14.74 \cdot 1.5 = 22.11 \mathrm{~m} \]- At \( t = 2.0 \mathrm{~s} \): \[ x(2.0) = 14.74 \cdot 2.0 = 29.48 \mathrm{~m} \]

Calculate the Y Position at Different Times

The vertical position \( y(t) \) is found using the equation \( y(t) = v_{y,0} \cdot t - \frac{1}{2}gt^2 \), where \( g = 9.8 \mathrm{~m/s^2} \). Calculate the y-position for each time:- At \( t = 0.5 \mathrm{~s} \): \[ y(0.5) = 10.32 \cdot 0.5 - \frac{1}{2} \cdot 9.8 \cdot 0.5^2 = 4.19 \mathrm{~m} \]- At \( t = 1.0 \mathrm{~s} \): \[ y(1.0) = 10.32 \cdot 1.0 - \frac{1}{2} \cdot 9.8 \cdot 1.0^2 = 5.42 \mathrm{~m} \]- At \( t = 1.5 \mathrm{~s} \): \[ y(1.5) = 10.32 \cdot 1.5 - \frac{1}{2} \cdot 9.8 \cdot 1.5^2 = 5.01 \mathrm{~m} \]- At \( t = 2.0 \mathrm{~s} \): \[ y(2.0) = 10.32 \cdot 2.0 - \frac{1}{2} \cdot 9.8 \cdot 2.0^2 = 2.96 \mathrm{~m} \]

Plot the Positions on Coordinate System

Using the calculated positions, plot the points \((x,y)\) on an \(x-y\) coordinate system:- \( (7.37, 4.19) \)- \( (14.74, 5.42) \)- \( (22.11, 5.01) \)- \( (29.48, 2.96) \)Connect these points with a smooth curve to form a parabolic trajectory, representing the path of the softball.

Key concepts

2D kinematics

2D kinematics is all about understanding the motion of an object in two dimensions, typically along the x and y axes. This type of analysis is crucial when studying projectiles, such as softballs thrown or launched at an angle. For projectiles, kinematics helps us predict the object's position, velocity, and acceleration at any given time.

• Position: Determining where the object is located at a specific time.
• Velocity: Understanding the speed and direction of the moving object.
• Acceleration: The rate of change of velocity, often influenced by gravity in projectile motion.

In our exercise, the softball's 2D motion has been analyzed by breaking its trajectory into horizontal (x) and vertical (y) components.
We can predict where the softball will land and how high it will travel over time by using these components.

Initial velocity components

When a softball is launched into the air, its initial speed is distributed into two components: the horizontal and the vertical. This distribution is influenced by the angle of projection, which determines the path and distance of the object.

To find the initial velocity components:

• Horizontal Component (\( v_{x,0} \)): This is calculated using the cosine of the launch angle. It represents the constant speed at which the object moves along the horizontal plane. In our exercise, this value is calculated as \( v_{x,0} = 18 \cdot \cos(35^{\circ}) \), resulting in approximately 14.74 m/s.
• Vertical Component (\( v_{y,0} \)): Found using the sine of the launch angle, this component determines how high and for how long the object will rise before gravity pulls it back down. For our softball, \( v_{y,0} = 18 \cdot \sin(35^{\circ}) \), equating to roughly 10.32 m/s.

Understanding these components helps predict the object's motion and determine the height and distance it will achieve over time.

Parabolic trajectory

Projectile motion typically follows a parabolic trajectory, which is a curved path characterized by the symmetric rise and fall of the projectile. This shape results from the constant horizontal velocity and the acceleration due to gravity acting downward.

As the softball in our problem is projected:

• The horizontal speed remains constant since there is no acceleration acting along the horizontal axis.
• The vertical motion is influenced by gravity, causing the softball to first rise to a peak and then descend, completing the parabola.

The trajectory is visible when the plotted positions from our exercise points, such as (7.37, 4.19) to (29.48, 2.96), form a smooth parabolic curve. This curve helps in visualizing the path and predicting the landing point of the projectile.

Coordinate system plotting

Plotting a projectile's motion on a coordinate system involves representing calculated position points on an x-y graph. This visual representation is essential for analyzing the projectile's path and understanding the motion dynamics.

To effectively plot the trajectory of a softball:

• Identify the x and y positions at various time intervals, like the 0.5s, 1.0s, 1.5s, and 2.0s marks used in our exercise.
• Plot these points on a grid, where the x-axis shows horizontal distance and the y-axis shows vertical height.

Once the points are plotted, connect them with a smooth curve to form a parabolic shape, clearly depicting the projectile's path.
This visual plot allows us to gauge the apex or peak height and predict points such as the landing location, crucial for real-world applications like sports or engineering.