Question

For a science fair competition, a group of high school students build a kicker-machine that can launch a golf ball from the origin with a velocity of \(11.2 \mathrm{~m} / \mathrm{s}\) and initial angle of \(31.5^{\circ}\) with respect to the horizontal. a) Where will the golf ball fall back to the ground? b) How high will it be at the highest point of its trajectory? c) What is the ball's velocity vector (in Cartesian components) at the highest point of its trajectory? d) What is the ball's acceleration vector (in Cartesian components) at the highest point of its trajectory?

Short answer

- The golf ball will land approximately 11.376 meters from the origin. b) What is the maximum height reached by the golf ball during its trajectory? - The golf ball will reach a maximum height of approximately 1.734 meters above the ground. c) What is the golf ball's velocity vector at the highest point of its trajectory? - The velocity vector at the highest point of the trajectory is approximately 𝑽 = ⟨9.514 m/s, 0 m/s⟩. d) What is the golf ball's acceleration vector at the highest point of its trajectory? - The acceleration vector at the highest point of the trajectory is ⟨0 m/s², -9.8 m/s²⟩.

Step-by-step solution

Find the time of flight

To determine where the golf ball falls back to the ground, we first need to find the time of flight. The time of flight is calculated using the vertical motion. The initial velocity in the vertical direction can be written as \(V_{0y} = V_0\sin(\theta)\), where \(V_0\) is the initial velocity and \(\theta\) is the launch angle. The time of flight can be found using the equation \(y = V_{0y}t - 0.5gt^2\), where \(y\) is the vertical displacement, \(t\) is the time, and \(g\) is the acceleration due to gravity (-9.8 \(\text{m}/\text{s}^2\)). At the time when the ball falls back to the ground, \(y = 0\). We can set up the equation as follows: $$0 = V_0\sin(\theta)t - 0.5gt^2$$

Solve for t

Now, we need to solve for \(t\) given the initial conditions. We know \(V_0 = 11.2\ \text{m}/\text{s}\) and \(\theta = 31.5^\circ\). First, find \(V_{0y}\): $$V_{0y} = 11.2\ \text{m}/\text{s}\sin(31.5^\circ)$$ $$V_{0y} \approx 5.862\ \text{m}/\text{s}$$ Plugging the values into the equation and solving for \(t\) (we'll consider only the positive value for time): $$ 0 = 5.862t - 4.9t^2$$ $$ t \approx 1.196\ \text{s}$$

Find the horizontal displacement

Using the time of flight, we can determine the horizontal displacement by using the initial velocity in the horizontal direction. The initial velocity in the horizontal direction can be written as \(V_{0x} = V_0\cos(\theta)\). We can use the equation \(x = V_{0x}t\) to find the horizontal displacement. First, find \(V_{0x}\): $$V_{0x} = 11.2\ \text{m}/\text{s}\cos(31.5^\circ)$$ $$V_{0x} \approx 9.514\ \text{m}/\text{s}$$ Now, we can find the horizontal displacement using the time of flight: $$x = 9.514\ \text{m}/\text{s} \times 1.196\ \text{s}$$ $$x \approx 11.376\ \text{m}$$ So, the golf ball will fall back to the ground at 𝟏𝟏.𝟑𝟕𝟔⁢𝒎 from the origin. #b) How high will it be at the highest point of its trajectory?#

Find the maximum height

The maximum height will occur when the vertical velocity component of the ball becomes zero. In other words, \(V_y = 0\). We can use the vertical motion equation to find this height: $$V_y^2 = V_{0y}^2 - 2gy$$ Setting \(V_y = 0\) and solving for \(y\): $$0 = 5.862^2 - 2(-9.8)y$$ $$y \approx 1.734\ \text{m}$$ Therefore, the highest point in its trajectory is at 1.734 meters above the ground. #c) What is the ball's velocity vector (in Cartesian components) at the highest point of its trajectory?#

Find the vertical component of velocity

At the highest point of its trajectory, the vertical component of velocity (\(V_{y})\) is 0.

Find the horizontal component of velocity

The horizontal component of velocity (\(V_{x})\) remains constant throughout the trajectory. Thus, at any point in time, including the highest point, the \(V_x \approx 9.514\ \text{m}/\text{s}\).

Write the velocity vector

Now, we can write the velocity vector at the highest point of its trajectory using both components: $$\vec{V} = \langle V_x, V_y \rangle = \langle 9.514\ \text{m}/\text{s}, 0\ \text{m}/\text{s} \rangle$$ The velocity vector at the highest point of the trajectory is approximately 𝑽=⟨9.514 m/s, 0 m/s⟩. #d) What is the ball's acceleration vector (in Cartesian components) at the highest point of its trajectory?#

Write the acceleration vector

Since there's no force applied to the ball during its trajectory except for gravitational force, the acceleration remains constant. The acceleration vector is given by: $$\vec{a} = \langle a_x, a_y \rangle = \langle 0\ \text{m}/\text{s}^2, -9.8\ \text{m}/\text{s}^2 \rangle$$ The acceleration vector at the highest point of the trajectory is ⟨0 m/s², -9.8 m/s²⟩.

Key concepts

Kinematic Equations

When studying projectile motion, we often use kinematic equations to describe the motion of the object in terms of its position, velocity, and acceleration over time. These equations are the tools that allow us to predict future movements from current measurements.

Specifically, in projectile motion, we have two components to consider: horizontal and vertical. The horizontal motion is uniform, meaning that there is no acceleration, while the vertical motion is uniformly accelerated due to gravity. The kinematic equations for vertical motion, which include terms for initial velocity (\( V_{0y} \text{or}\ V_0\text{sin}\theta \text{where}\theta \text{is the launch angle} \)), time (\( t \text{how long the projectile is in motion} \)), acceleration due to gravity (\( g \)), and vertical position (\( y \)), provide the basis for solving many problems related to projectile motion.

Trajectory of a Projectile

The trajectory of a projectile is the path it follows under the influence of gravity alone. After being projected, the object follows a curved path known as a parabola before returning to the ground. This parabolic trajectory is influenced by the initial speed and the angle of launch.

With an angle of elevation of \(31.5^\text{\circ}\) and an initial speed, our golf ball moves out and up, then down and out, tracing a symmetric path. The trajectory is determined by both the vertical and horizontal components of motion, with gravity affecting only the vertical component. The analysis of projectile trajectory allows us to calculate the maximum height the object reaches and the distance it travels horizontally before touching the ground.

Time of Flight

The time of flight in projectile motion refers to the total duration the projectile remains airborne. For our golf ball, this is the time from launch until it returns to the ground. It is directly related to the vertical component of the initial velocity and the acceleration due to gravity.

To calculate the time of flight, we consider the moment when the golf ball lands (\( y = 0 \)) and solve for \( t \) considering upward motion (positive) and downward motion (negative due to gravity). The equation \( 0 = V_0\text{sin}(\theta)t - 0.5gt^2 \) helps us find that time, and once it's known, it allows us to calculate other properties, like the horizontal displacement of the projectile.

Maximum Height of Projectile

The maximum height a projectile achieves occurs at the apex of its trajectory, where the vertical component of velocity momentarily becomes zero (\( V_{y} = 0 \)). At this point, the only component of velocity that remains is the horizontal one, as it is unaffected by gravity.

Using the equation \( V_{y}^2 = V_{0y}^2 - 2gy \), and setting \( V_{y} = 0 \) helps to isolate for \( y \) giving us the maximum height. This equation reveals the direct relationship between the initial speed of the projectile, the angle of projection, and the peak height it can reach in the absence of any other forces like air resistance.

Velocity Vector Components

The velocity of a projectile is composed of both horizontal and vertical velocity vector components. At launch, the velocity will have both a positive vertical component and a horizontal component, based on the projection angle. However, at the peak of its flight, the vertical component becomes zero (\( V_{y} = 0 \)) while the horizontal component remains constant due to the absence of horizontal forces in ideal projectile motion.

The constant horizontal velocity component (\( V_{0x} \)) can be calculated using \( V_{0x} = V_0\text{cos}(\theta) \). These vector components allow us to understand the motion of the projectile in more detail at any point in its trajectory.

Acceleration Due to Gravity

The acceleration due to gravity is a constant value (\(9.8 \text{m/s}^2\) on the surface of Earth) that acts on all objects equally, pulling them towards the Earth's center. This acceleration is always directed downward in vertical projectile motion problems.

In the absence of air resistance, the only acceleration affecting our golf ball is the acceleration due to gravity, hence why its vertical component of acceleration is \(-9.8 \text{m/s}^2\), which is constant throughout the motion, while the horizontal component of acceleration is zero.