Projectile Motion
Textbook Reference
| Primary source | OpenStax Calculus Volume 3, Section 3.4: “Motion in Space” |
| Direct link | https://openstax.org/books/calculus-volume-3/pages/3-4-motion-in-space |
| Textbook used in class | Stewart, Calculus, Section 13.4: “Motion in Space: Velocity and Acceleration” (Example 3) |
Quick Reference
Ideal projectile (gravity $g = 9.8$ m/s$^2$ or $32$ ft/s$^2$, no air resistance): $$\mathbf{a}(t) = \langle 0, -g \rangle \quad \text{(2D)}, \qquad \mathbf{a}(t) = \langle 0, -g, 0 \rangle \quad \text{(3D)}.$$
Position (launched from origin with speed $v_0$ at angle $\alpha$): $$x(t) = v_0 \cos\alpha\cdot t, \qquad y(t) = v_0\sin\alpha\cdot t - \tfrac{1}{2}g t^2.$$
Maximum height: $y_{\max} = \dfrac{v_0^2 \sin^2\alpha}{2g}$ (attained when $v_y = 0$).
Range: $R = \dfrac{v_0^2 \sin 2\alpha}{g}$ (horizontal distance when the projectile returns to $y = 0$).
Motivation
Projectile motion is the simplest application of the vector kinematics from the previous skill: the only force acting is gravity, which produces a constant downward acceleration. Despite this simplicity, the resulting parabolic path describes the flight of thrown objects, artillery shells, and water jets.
The vector approach integrates the constant acceleration directly to get position as a function of time, making maximum height and range immediate consequences of solving for when $v_y(t) = 0$ and $y(t) = 0$.
Key Concepts
1. Setting Up the Model
In the ideal model, air resistance is neglected and gravity acts downward at $g$ m/s$^2$. Taking the $y$-axis as vertical: $$\mathbf{a}(t) = \langle 0, -g\rangle.$$
Integrating once gives velocity, integrating again gives position. Each integration introduces a vector constant determined by initial conditions.
2. The Parametric Path
For a projectile launched from the origin with initial speed $v_0$ at angle $\alpha$ above the horizontal: $$\mathbf{v}(0) = \langle v_0\cos\alpha,\; v_0\sin\alpha\rangle.$$
After integration: $$\mathbf{v}(t) = \langle v_0\cos\alpha,\; v_0\sin\alpha - gt\rangle,$$ $$\mathbf{r}(t) = \langle v_0\cos\alpha\cdot t,\; v_0\sin\alpha\cdot t - \tfrac{1}{2}gt^2\rangle.$$
Eliminating $t$ from $x = v_0\cos\alpha \cdot t$ gives a parabola $y = x\tan\alpha - \dfrac{g x^2}{2v_0^2\cos^2\alpha}$, confirming the path is parabolic.
3. Maximum Height and Range
Max height: The projectile reaches maximum height when the vertical velocity is zero: $$v_0\sin\alpha - gt^* = 0 \implies t^* = \frac{v_0\sin\alpha}{g}.$$ Substituting: $y_{\max} = \dfrac{v_0^2\sin^2\alpha}{2g}$.
Range: The projectile lands when $y(t) = 0$ again (and $t > 0$): $$t_{\text{land}} = \frac{2v_0\sin\alpha}{g}.$$ $$R = x(t_{\text{land}}) = v_0\cos\alpha \cdot \frac{2v_0\sin\alpha}{g} = \frac{v_0^2\sin 2\alpha}{g}.$$
The range is maximized when $\sin 2\alpha = 1$, i.e., $\alpha = 45°$.
Worked Example
A ball is thrown horizontally from a height of 20 m with speed 15 m/s. Find when and where it hits the ground. ($g = 9.8$ m/s$^2$.)
Setup. Place the origin at the launch point. The initial conditions are $\mathbf{r}(0) = \langle 0, 0\rangle$ (measuring height from ground: the ball starts at $y = 0$ in our shifted frame, and the ground is at $y = -20$), or equivalently: set the origin at ground level and $\mathbf{r}(0) = \langle 0, 20\rangle$.
Using $\mathbf{r}(0) = \langle 0, 20\rangle$ and $\mathbf{v}(0) = \langle 15, 0\rangle$ (horizontal throw means no vertical initial speed):
$$\mathbf{a}(t) = \langle 0, -9.8\rangle.$$ $$\mathbf{v}(t) = \langle 15, -9.8t\rangle.$$ $$\mathbf{r}(t) = \langle 15t,\; 20 - 4.9t^2\rangle.$$
Landing: $y(t) = 0$ gives $20 - 4.9t^2 = 0$, so $t = \sqrt{20/4.9} \approx 2.02$ s.
Horizontal distance: $x = 15(2.02) \approx 30.3$ m.
The ball hits the ground approximately 30.3 m from the base of the launch point after about 2.02 seconds.
a horizontally thrown ball falls slower because it is moving forward. The horizontal and vertical components of motion are completely independent in the ideal model. Gravity accelerates the ball downward at exactly the same rate regardless of horizontal speed. A ball thrown horizontally and one dropped from the same height hit the ground at the same time.
Common Misconceptions
horizontal speed slows the vertical fall of a projectile.
This is the composition-is-not-chaining error. In the ideal model, horizontal and vertical motions are completely independent. Gravity acts only on the vertical component; the horizontal component has zero acceleration and maintains its initial speed. A ball thrown horizontally and a ball dropped from rest at the same height fall at exactly the same rate and hit the ground at the same time. The horizontal distance covered while falling does not affect the time to reach the ground.
Leveled Practice
Problem 1. A projectile is launched from the ground with $v_0 = 50$ m/s at $\alpha = 30°$. Find the maximum height and the range. ($g = 9.8$ m/s$^2$, $\sin 30° = 0.5$, $\cos 30° = \sqrt{3}/2 \approx 0.866$.)
Show answer
$v_{0x} = 50(0.866) = 43.3$ m/s, $v_{0y} = 50(0.5) = 25$ m/s.
$y_{\max} = \dfrac{25^2}{2(9.8)} = \dfrac{625}{19.6} \approx 31.9$ m.
$R = \dfrac{50^2 \sin 60°}{9.8} = \dfrac{2500(0.866)}{9.8} \approx 221$ m.