2.1 - Projectile Motion in a Vacuum

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Introduction

Before we discuss projectile motion with air drag, let's review projectile motion in a vacuum. Projectile motion is a common problem in first-year physics courses. When an object is launched with an initial velocity and allowed to move under the influence of gravity alone (neglecting air resistance), its motion follows a parabolic trajectory. This problem provides an introduction to solving two-dimensional motion by decomposing the equations into independent $x$ and $y$ components.

Force Components:

For projectile motion in the absence of air drag, the force components are:

  • $x$-direction: $F_x = 0$ (no horizontal force)
  • $y$-direction: $F_y = -mg$ (gravitational force, downward)

where $m$ is the mass of the object and $g$ is the acceleration due to gravity (approximately $9.8$ m/s² near Earth's surface).

Setting Up the Problem

Consider a projectile launched from the origin with initial velocity $\mathbf{v}_0$ at an angle $\theta$ above the horizontal. The initial conditions are:

where $v_0 = |\mathbf{v}_0|$ is the initial speed.

Solving in the $x$-Direction

Becuase the net force in the $x$-direction is zero, Newton's Second Law $F_x = ma_x$ gives:

$$\frac{dv_x}{dt} = 0.$$

This means that the horizontal velocity is constant and equal to its initial value:

Horizontal Velocity:
$$v_x = v_0\cos\theta.$$

The horizontal velocity is constant throughout the motion.

Now, to find the horizontal position, we use the relationship $v_x = dx/dt$:

$$\frac{dx}{dt} = v_0\cos\theta$$

Cross-multiplying by $dt$ and integrating both sides gives the following result.

Horizontal Position:
$$x(t) = [v_0\cos\theta] t.$$

Solving in the $y$-Direction

Starting with Newton's Second Law in the $y$-direction

$$F_y = ma_y,$$

we substitute $F_y = -mg$ and $a_y = \frac{dv_y}{dt}$ to get:

$$-mg = m\frac{dv_y}{dt}$$

Divide by $m$

$$\frac{dv_y}{dt} = -g,$$

cross-multiply by dt and integrating both sides:

$$\int_{v_{0y}}^{v_y} dv_y = \int_0^t -g \, dt.$$

This gives:

$$v_y - v_{0y} = -gt.$$

Substituting the initial condition $v_{0y} = v_0\sin\theta$ gives th final result.

Vertical Velocity:
$$v_y(t) = v_0\sin\theta - gt$$

Now, to find the vertical position, we use the relationship $v_y = dy/dt$:

$$\frac{dy}{dt} = v_0\sin\theta - gt.$$

Integrate both sides:

$$\int_{u_0}^y dy = \int_0^t (v_0\sin\theta - gt) \, dt$$

gives the final result.

Vertical Position:
$$y(t) = v_0\sin\theta \cdot t - \frac{1}{2}gt^2$$

Summary: Parametric Equations of Motion

The complete solution for projectile motion without air drag is given by the parametric equations:

Projectile Motion Equations:
$$x(t) = v_0\cos\theta \cdot t$$
$$y(t) = v_0\sin\theta \cdot t - \frac{1}{2}gt^2$$

where $v_0$ is the initial speed, $\theta$ is the launch angle, and $g$ is the acceleration due to gravity.

Properties of the Trajectory

From these equations, we can derive several important results:

Time of Flight

The time of flight is found by setting $y(t) = 0$ (projectile returns to ground level):

$$y(t) = v_0\sin\theta \cdot t - \frac{1}{2}gt^2 = 0.$$

Factor out $t$

$$t\left(v_0\sin\theta - \frac{1}{2}gt\right) = 0.$$

The solutions are $t = 0$ (launch) and:

$$t_\text{flight} = \frac{2v_0\sin\theta}{g}.$$

This is the total time of flight.

Maximum Height

The maximum height occurs when $v_y = 0$:

$$v_y(t) = v_0\sin\theta - gt = 0 \quad \Rightarrow \quad t_\text{top} = \frac{v_0\sin\theta}{g}.$$

Substituting $t_\text{top}$ into $y(t)$ and calling this height $y_\text{max}$:

$$\begin{aligned} y(t) &= v_0\sin\theta \cdot t - \frac{1}{2}gt^2\\ &= v_0\sin\theta \left(\frac{v_0\sin\theta}{g}\right) - \frac{1}{2}g\left(\frac{v_0\sin\theta}{g}\right)^2\\ &= \frac{v_0^2\sin^2\theta}{g} - \frac{v_0^2\sin^2\theta}{2g}. \end{aligned}$$

This gives the maximum height:

$$y_\text{max} = \frac{v_0^2\sin^2\theta}{2g}.$$

Range

The horizontal range is found by substituting the time of flight $t = 2v_0\sin\theta/g$ into $x(t) = v_0\cos\theta \cdot t$:

$$R = v_0\cos\theta \left( \frac{2v_0\sin\theta}{g} \right) = \frac{2v_0^2\sin\theta\cos\theta}{g}.$$

Using the trigonometric identity $\sin(2\theta) = 2\sin\theta\cos\theta$ gives the desired expression for the range:

$$R = \frac{v_0^2\sin(2\theta)}{g}.$$

The maximum range occurs when $\sin(2\theta) = 1$, i.e., when $\theta = 45°$.

Trajectory Shape:

By eliminating $t$ from the parametric equations, we can find the trajectory shape. From $x(t) = v_0\cos\theta \cdot t$, we have $t = x/(v_0\cos\theta)$. Substituting into $y(t)$ gives:

$$y = v_0\sin\theta \cdot \frac{x}{v_0\cos\theta} - \frac{1}{2}g\left(\frac{x}{v_0\cos\theta}\right)^2.$$

This is the equation of a parabola, confirming that projectile motion follows a parabolic trajectory.

$$y = \left[ \tan\theta \right] x- \left[ \frac{g}{2v_0^2\cos^2\theta} \right] x^2$$

Interactive Trajectory Plot

Use the interactive plot below to explore how the initial speed and launch angle affect the projectile trajectory. Adjust the sliders to see how the trajectory changes in real time.

Interactive Projectile Motion

50.0 m/s
45°
Range: 255.10 m | Maximum Height: 63.78 m | Time of Flight: 7.22 s