Intro to Projectile Motion: Horizontal Launch: Videos & Practice Problems

Projectile motion involves the movement of an object in two dimensions under the influence of gravity, resulting in a parabolic path. Key concepts include decomposing motion into horizontal and vertical components, where horizontal motion has zero acceleration and vertical motion experiences acceleration due to gravity (g). To solve problems, identify the target variable, determine the relevant intervals, and apply appropriate equations, such as t2= 2 - Δy / (-g). Understanding these principles is essential for analyzing real-world scenarios in physics.

Introduction to Projectile Motion

Projectile motion is a fundamental concept in physics that describes the motion of an object thrown into the air, moving in two dimensions under the influence of gravity. Unlike one-dimensional vertical motion, where an object moves straight up and down, projectile motion involves an initial sideways velocity, resulting in a parabolic trajectory. This motion can occur in various forms, such as launching an object horizontally, throwing it downward, or tossing it upward. Regardless of the specific scenario, the key factor is that the only force acting on the object is gravity, which accelerates it downward at approximately 9.81 m/s².

To analyze projectile motion, it is essential to decompose the motion into its horizontal (x-axis) and vertical (y-axis) components. This approach allows us to treat the two dimensions separately. For instance, if an object is thrown from point A to point B, its horizontal motion can be viewed as a constant velocity motion, while its vertical motion is influenced by gravity. The horizontal acceleration is zero, meaning the object moves uniformly along the x-axis, while the vertical motion experiences a constant acceleration of -g (where g is the acceleration due to gravity).

In terms of equations, the horizontal motion can be described using the formula for constant velocity:

v_x = \frac{d_x}{t}

where v_x is the horizontal velocity, d_x is the horizontal distance traveled, and t is the time of flight. In contrast, the vertical motion is governed by the equations of motion for uniformly accelerated motion, which can be expressed as:

d_y = v_{y0}t + \frac{1}{2}(-g)t^2

where d_y is the vertical displacement, v_{y0} is the initial vertical velocity, and t is the time of flight.

By applying these principles, students can solve a variety of projectile motion problems, recognizing that the combination of horizontal and vertical motions results in the characteristic parabolic path. Understanding these concepts is crucial for tackling more complex physics problems involving two-dimensional motion.

Which of the following quantities are constant during projectile motion?

• A. Vertical acceleration & vertical velocity
• B. Angle (direction) of the velocity vector
• C. Horizontal acceleration & vertical velocity
• D. Vertical acceleration & horizontal velocity

Answer: (D)

Solving Horizontal Launch Problems

Projectile motion involves the movement of an object under the influence of gravity, typically characterized by a parabolic trajectory. To analyze such motion, especially in cases of horizontal launches, a systematic approach can be employed. This involves breaking down the motion into horizontal (x-axis) and vertical (y-axis) components.

When a ball rolls off a table, the first step is to sketch the trajectory on the x and y axes. The ball will follow a curved path due to gravity, and it is essential to identify key points, such as the initial position (point A) and the final position (point B) when the ball hits the ground. The next step is to determine the target variable, which in this case is the time (t) it takes for the ball to reach the ground.

To find the time of flight, we can use the equation for horizontal motion:

$$\Delta x = v_{x} \cdot t_{AB}$$

where $$\Delta x$$ is the horizontal displacement, $$v_{x}$$ is the horizontal velocity, and $$t_{AB}$$ is the time from point A to point B. In horizontal launches, the initial velocity is entirely in the x-direction, and the vertical component of the initial velocity is zero.

For the vertical motion, we can use the following kinematic equation, which relates displacement, initial velocity, acceleration, and time:

$$\Delta y = v_{yA} \cdot t_{AB} + \frac{1}{2} a_{y} \cdot t_{AB}^2$$

In this equation, $$\Delta y$$ is the vertical displacement (which is negative if the object falls), $$v_{yA}$$ is the initial vertical velocity (zero for horizontal launches), and $$a_{y}$$ is the acceleration due to gravity (approximately -9.8 m/s²).

For example, if the table height is 2 meters, the vertical displacement is -2 meters. Plugging in the known values allows us to solve for time:

$$-2 = 0 \cdot t_{AB} + \frac{1}{2} (-9.8) \cdot t_{AB}^2$$

Solving this gives us the time of flight, which is approximately 0.64 seconds.

Once the time is determined, we can find the horizontal displacement using the previously mentioned equation. If the horizontal velocity is 3 m/s, the horizontal displacement can be calculated as:

$$\Delta x = 3 \cdot 0.64 = 1.92 \text{ meters}$$

In summary, for horizontal projectile motion, the initial velocity is entirely in the x-direction, and the vertical motion is influenced solely by gravity. The time of flight can be calculated using vertical motion equations, while horizontal displacement can be determined using the constant horizontal velocity. Understanding these principles allows for effective problem-solving in projectile motion scenarios.

A rock is thrown horizontally with a speed of 20 m/s from the edge of a high cliff. It lands 80 m from the cliff's base. How tall is the cliff?

• A. 78.4 m
• B. 19.6 m
• C. 122.5 m
• D. 24.5 m

Answer: (A)

Clearing the Net

In this problem, we are tasked with determining the initial speed required for a ping pong ball to clear a net when served horizontally. The net is positioned 1.6 meters away from the player and is 0.9 meters high. To solve this, we need to analyze the motion of the ball in both the horizontal (x) and vertical (y) directions.

First, we establish the coordinates for our points of interest: point A (the initial position of the ball) and point B (the point where the ball must clear the net). The horizontal distance, or displacement, from A to B is given as Δx = 1.6 meters. The height of the net above the table is 0.9 meters, and the ball is served from a height of 1.2 meters, resulting in a vertical displacement of Δy = -0.3 meters (the negative sign indicates a downward movement).

Since the ball is served horizontally, its initial vertical velocity (v₀y) is 0. The acceleration due to gravity (aₓ) acting on the ball is -9.8 m/s². To find the time (t) it takes for the ball to reach the net, we can use the kinematic equation that relates vertical displacement to time:

$$\Delta y = v_{0y} t + \frac{1}{2} a_y t^2$$

Substituting the known values, we have:

$$-0.3 = 0 \cdot t + \frac{1}{2} (-9.8) t^2$$

This simplifies to:

$$-0.3 = -4.9 t^2$$

Solving for t gives:

$$t^2 = \frac{0.3}{4.9} \implies t \approx 0.25 \text{ seconds}$$

Now that we have the time, we can return to the horizontal motion to find the initial horizontal velocity (v₀x). The horizontal motion is described by the equation:

$$\Delta x = v_{0x} t$$

Substituting the known values:

$$1.6 = v_{0x} \cdot 0.25$$

Solving for v₀x gives:

$$v_{0x} = \frac{1.6}{0.25} = 6.4 \text{ m/s}$$

Thus, the ping pong ball must be served with an initial speed of 6.4 meters per second to successfully clear the net. This calculation highlights the importance of understanding projectile motion, particularly the separation of horizontal and vertical components of motion, and how they interact to determine the trajectory of an object.

An Unfortunate Landing

In this problem, we analyze a scenario where a ball is kicked horizontally from the top of a building, and a car accelerates from rest to catch the ball as it falls. The goal is to determine the car's acceleration required to meet the ball at the ground.

To solve this, we first establish the position equations for both the ball and the car. The ball follows a parabolic trajectory due to its horizontal launch, while the car accelerates from rest. The equations for their positions can be expressed as:

For the ball:

\( x_{\text{ball}} = v_{\text{ball}} \cdot t \)

For the car:

\( x_{\text{car}} = \frac{1}{2} a_{\text{car}} \cdot t^2 \)

Here, \( v_{\text{ball}} \) is the initial horizontal velocity of the ball, and \( a_{\text{car}} \) is the acceleration of the car, which we are trying to find. Since the ball is kicked horizontally, its vertical motion is influenced by gravity, with an acceleration of \( g = 9.8 \, \text{m/s}^2 \).

Next, we set the two position equations equal to each other to find the time \( t \) when both the ball and the car are at the same horizontal position:

\( v_{\text{ball}} \cdot t = \frac{1}{2} a_{\text{car}} \cdot t^2 \)

By simplifying, we can cancel \( t \) (assuming \( t \neq 0 \)) to get:

\( v_{\text{ball}} = \frac{1}{2} a_{\text{car}} \cdot t \)

To find \( t \), we analyze the vertical motion of the ball. The vertical displacement \( \Delta y \) from the top of the building to the ground is \( -40 \, \text{m} \) (downward). Using the kinematic equation that relates displacement, initial velocity, acceleration, and time:

\( \Delta y = v_{ay} \cdot t + \frac{1}{2} a_y \cdot t^2 \)

Since the initial vertical velocity \( v_{ay} = 0 \), the equation simplifies to:

\( -40 = \frac{1}{2} (-9.8) \cdot t^2 \

Solving for \( t \), we rearrange to find:

\( t^2 = \frac{2 \cdot 40}{9.8} \)

\( t = \sqrt{\frac{80}{9.8}} \approx 2.86 \, \text{s} \)

Now that we have \( t \), we can substitute it back into the equation for the ball's horizontal motion:

\( 8 = \frac{1}{2} a_{\text{car}} \cdot 2.86 \)

Rearranging gives:

\( a_{\text{car}} = \frac{8 \cdot 2}{2.86} \approx 5.60 \, \text{m/s}^2 \)

This means the car must accelerate at approximately \( 5.60 \, \text{m/s}^2 \) to catch the ball as it falls. This problem illustrates the integration of projectile motion and kinematics, highlighting the importance of understanding both horizontal and vertical components of motion in physics.

Here’s what students ask on this topic:

What is projectile motion and how is it different from vertical motion?

Projectile motion involves the movement of an object in two dimensions under the influence of gravity, resulting in a parabolic path. Unlike vertical motion, which occurs in one dimension (up and down), projectile motion has both horizontal and vertical components. The horizontal motion has zero acceleration, meaning the horizontal velocity remains constant. In contrast, the vertical motion experiences acceleration due to gravity (g), which affects the vertical velocity. This combination of constant horizontal velocity and accelerating vertical motion creates the characteristic parabolic trajectory of projectile motion.

How do you decompose projectile motion into horizontal and vertical components?

To decompose projectile motion into horizontal and vertical components, you analyze the motion separately along the x-axis (horizontal) and y-axis (vertical). For the horizontal component, the acceleration is zero, so the velocity remains constant. The equation used is $\mathrm{\Delta x}=v$_xt. For the vertical component, the motion is influenced by gravity, with an acceleration of $a=-g$. The equations of motion for the vertical component include $\mathrm{\Delta y}=v$_y0t+12at². By solving these equations separately, you can determine the overall motion of the projectile.

What are the key equations used to solve projectile motion problems?

The key equations for solving projectile motion problems are divided into horizontal and vertical components. For horizontal motion, the primary equation is $\mathrm{\Delta x}=v$_xt, where $v$_x is the constant horizontal velocity. For vertical motion, the equations include: $\mathrm{\Delta y}=v$_y0t+12at², $v$_y=v_y0+at, and $v$_y²=v_y0²+2aΔy, where $a=-g$ (acceleration due to gravity). These equations help determine the time of flight, range, and maximum height of the projectile.

How do you calculate the time of flight for a horizontally launched projectile?

To calculate the time of flight for a horizontally launched projectile, you focus on the vertical motion. The time of flight depends on the height from which the projectile is launched. Using the equation $\mathrm{\Delta y}=\frac{1}{2}gt2$, where $\mathrm{\Delta y}$ is the vertical displacement (negative if falling), and $g$ is the acceleration due to gravity. Rearrange to solve for time: $t=\sqrt{\frac{2\mathrm{\Delta y}}{g}}$. Plug in the values for $\mathrm{\Delta y}$ and $g$ to find the time of flight.

What is the significance of zero acceleration in the horizontal component of projectile motion?

The significance of zero acceleration in the horizontal component of projectile motion is that the horizontal velocity remains constant throughout the motion. This simplifies calculations because the horizontal displacement can be found using the equation $\mathrm{\Delta x}=v$_xt, where $v$_x is the initial horizontal velocity and $t$ is the time of flight. Since there is no horizontal acceleration, the horizontal motion is uniform, making it easier to analyze and predict the projectile's path. This concept is crucial for solving projectile motion problems accurately.