Lesson 5.3: Horizontal Range and Trajectory

Introduction

In this lesson, we will explore the world of projectiles, focusing specifically on their horizontal range and trajectory. Projectiles are fascinating because they exhibit a unique two-dimensional motion under the influence of gravity. Understanding this motion is essential for various applications, from sports to engineering. By the end of this lesson, you will be able to:

Find the horizontal range from the time of flight and the horizontal component of velocity.
Describe the position of a projectile at a given time as a vector or in terms of x and y coordinates.
Derive the equation of the trajectory and determine where a projectile lands.
Calculate the horizontal range of a projectile on level ground.
Find the position of a projectile at a given time and determine when it reaches a specified height.

Understanding Horizontal Motion

Concept of Horizontal Motion

When dealing with projectiles, we analyze the motion in two dimensions: horizontal and vertical. In this section, we will focus on horizontal motion, which is characterized by constant velocity, as there are no external horizontal forces acting on the projectile (assuming air resistance is negligible).

The horizontal motion can be described using the formula:

$$x = v_{x} \cdot t$$

where:

$x$ is the horizontal displacement,
$v_{x}$ is the horizontal component of velocity,
$t$ is the time of flight.

Worked Example 1: Horizontal Displacement Calculation

Problem: A projectile is launched with a horizontal component of velocity $v_{x} = 10 \, \text{m/s}$. Calculate the horizontal displacement after $t = 5 \, \text{s}$.

Solution: Using the horizontal motion formula:

$$x = v_{x} \cdot t = 10 \, \text{m/s} \cdot 5 \, \text{s} = 50 \, \text{m}$$

Hence, the horizontal displacement after 5 seconds is 50 meters.

Common Misconceptions

One common misconception is that horizontal motion under gravity is affected by gravitational force. However, horizontal motion is independent of vertical motion, and hence it continues at a constant velocity unless acted upon by another force.

Time of Flight

Deriving Time of Flight

The time of flight is crucial in projectile motion calculations. It can be derived from vertical motion, where the vertical distance is affected by gravity. The formula for the vertical motion of a projectile launched with an initial vertical velocity $v_{y}$ is:

$$y = v_{y} \cdot t - \frac{1}{2}gt^{2}$$

where:

$y$ is the vertical height,
$g$ is the acceleration due to gravity (approximately $9.81 \, \text{m/s}^{2}$).

Worked Example 2: Calculating Time of Flight

Problem: A projectile is launched vertically with an initial vertical velocity of $v_{y} = 20 \, \text{m/s}$ and lands at ground level ($y = 0$). Calculate the time of flight.

Solution: Setting the vertical motion equation to zero:

$$0 = 20 \cdot t - \frac{1}{2} \cdot 9.81 t^{2}$$

This simplifies to:

$$0 = 20t - 4.905t^{2}$$

Factoring out $t$ gives:

$$t(20 - 4.905t) = 0$$

Thus, $ t = 0 $ (initial launch) or $ t = \frac{20}{4.905} \approx 4.08 \, \text{s} $. Therefore, the time of flight is approximately 4.08 seconds.

Horizontal Range Calculation

Finding the Horizontal Range

The horizontal range can be calculated using the horizontal component of velocity and the time of flight. The formula is:

$$R = v_{x} \cdot T$$

where:

$R$ is the horizontal range,
$v_{x}$ is the horizontal component of velocity,
$T$ is the time of flight.

Worked Example 3: Calculating the Horizontal Range

Problem: A projectile is launched with a horizontal component of velocity $v_{x} = 15 \, \text{m/s}$ and a time of flight of $T = 3 \, \text{s}$. Find the horizontal range.

Solution: Using the range formula:

$$R = v_{x} \cdot T = 15 \cdot 3 = 45 \, \text{m}$$

Thus, the horizontal range of the projectile is 45 meters.

Projectile Position as a Vector

Representing Position Vectors

The position of a projectile can be represented in two ways: as a vector or as coordinates. The position vector, $\mathbf{r}(t)$, at time $t$ can be expressed as:

$$\mathbf{r}(t) = \begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} v_{x} \cdot t \ v_{y} \cdot t - \frac{1}{2}gt^{2} \end{pmatrix}$$

where:

$x$ is the horizontal position,
$y$ is the vertical position.

Worked Example 4: Position Vector Calculation

Problem: A projectile has a horizontal component of velocity $v_{x} = 12 \, \text{m/s}$ and an initial vertical velocity $v_{y} = 16 \, \text{m/s}$. Calculate the position vector at $t = 2 \, \text{s}$.

Solution: First, find the horizontal and vertical positions:

Horizontal position:

$$x = v_{x} \cdot t = 12 \cdot 2 = 24 \, \text{m}$$

Vertical position:

$$y = v_{y} \cdot t - \frac{1}{2}gt^{2} = 16 \cdot 2 - \frac{1}{2} \cdot 9.81 \cdot (2)^2 = 32 - 19.62 = 12.38 \, \text{m}$$

Thus, the position vector is:

$$\mathbf{r}(2) = \begin{pmatrix} 24 \ 12.38 \end{pmatrix}$$

Deriving the Equation of the Trajectory

Equation of the Trajectory

The trajectory of a projectile is represented in the form of a quadratic equation that relates $x$ and $y$. This is derived from the horizontal and vertical motion equations, leading to:

$$y = x \tan(\theta) - \frac{g}{2v_{x}^{2}} x^{2}$$

where:

$\theta$ is the angle of launch.

Worked Example 5: Trajectory Equation Calculation

Problem: A projectile is launched at an angle of $30^{\circ}$ with an initial speed of $v = 20 \, \text{m/s}$. Determine the equation of the trajectory.

Solution: First, find the horizontal and vertical component of velocity:

$v_{x} = v \cos(\theta) = 20 \cos(30^{\circ}) = 20 \cdot \frac{\sqrt{3}}{2} = 10\sqrt{3} \approx 17.32 \, \text{m/s}$
$v_{y} = v \sin(\theta) = 20 \sin(30^{\circ}) = 20 \cdot \frac{1}{2} = 10 \, \text{m/s}$

Substituting into the trajectory equation:

$$y = x \tan(30^{\circ}) - \frac{9.81}{2(10\sqrt{3})^{2}} x^{2}$$

Calculating the second term:

$$y = x \cdot \frac{1}{\sqrt{3}} - \frac{9.81}{200} x^{2} \approx x \cdot 0.577 - 0.04905 x^{2}$$

This gives us the equation of the trajectory:

$$y \approx 0.577x - 0.04905x^{2}$$

Conclusion

In this lesson, we have covered essential topics related to the horizontal range and trajectory of projectiles. We have analyzed how to calculate horizontal range using time of flight, represented the position of projectiles in vector form, and derived the equation of a trajectory. These concepts are fundamental in understanding the motion of projectiles and have practical applications in various fields such as sports, engineering, and physics.

Study Notes

Horizontal motion is characterized by constant velocity due to negligible horizontal forces.
Time of flight can be calculated from vertical motion equations.
Horizontal range is found using $R = v_{x} \cdot T$.
Position vectors represent projectile positions at given times $\mathbf{r}(t) = \begin{pmatrix} v_{x} \cdot t \ v_{y} \cdot t - \frac{1}{2}gt^{2} \end{pmatrix}$.
The equation of trajectory can be expressed as a quadratic function of horizontal displacement.