A projectile is any object moving only under gravity (no air drag). Examples: soccer ball, arrow, water fountain, artillery shell. The classic application of §4.1 and §4.2.
Fundamental equations
Set x horizontal, y vertical (up positive). Acceleration only downward:
\[ \vec a = -g\hat j = 0\hat i - 9.8\hat j\ \mathrm{m/s^2} \]
Initial velocity of magnitude v_0 at angle \theta_0 above horizontal:
\[ v_{0x} = v_0 \cos\theta_0,\quad v_{0y} = v_0 \sin\theta_0 \]
Full equations:
\[ x(t) = v_{0x}\, t \]
\[ y(t) = v_{0y}\, t - \frac{1}{2} g t^2 \]
\[ v_x(t) = v_{0x}\ (\text{constant!}) \]
\[ v_y(t) = v_{0y} - g t \]
Principle of independent motions
Horizontal: uniform — v_x never changes.
Vertical: free fall — like §2.5.
Time to peak, peak height
Time to peak (when v_y = 0):
\[ t_\text{peak} = \frac{v_{0y}}{g} = \frac{v_0 \sin\theta_0}{g} \]
Peak height:
\[ y_\text{max} = \frac{v_{0y}^2}{2g} = \frac{v_0^2 \sin^2\theta_0}{2g} \]
Total flight time and range
If the projectile lands at the same height as launch (y = 0 at end):
Total flight time:
\[ T = \frac{2 v_{0y}}{g} = \frac{2 v_0 \sin\theta_0}{g} \]
Horizontal range:
\[ R = v_{0x}\, T = \frac{v_0^2 \sin(2\theta_0)}{g} \]
Important result: range is maximum when \sin(2\theta_0) = 1, i.e., 2\theta_0 = 90°:
\[ \boxed{\theta_0 = 45°\ \text{for maximum range}} \]
Trajectory equation
Eliminate t between x(t) and y(t):
\[ y = x \tan\theta_0 - \frac{g x^2}{2 v_0^2 \cos^2\theta_0} \]
This is a parabola — the shape you actually see.
Complete example — baseball throw
Scenario: ball launched at v_0 = 30 m/s at \theta_0 = 40° from height y_0 = 1 m. Landing time, distance, impact speed?
Decompose:
v_{0x} = 30 \cos 40° \approx 22.98\ m/sv_{0y} = 30 \sin 40° \approx 19.28\ m/s
Landing at y = 0:
\[ 0 = 1 + 19.28 t - 4.9 t^2 \]
Quadratic:
\[ t = \frac{19.28 + \sqrt{19.28^2 + 4 \cdot 4.9}}{9.8} \approx 3.99\ s \]
Horizontal range: x = 22.98 \cdot 3.99 \approx 91.7\ m
Impact velocity:
v_x = 22.98\ m/sv_y = 19.28 - 9.8 \cdot 3.99 \approx -19.82\ m/s- Speed:
\sqrt{22.98^2 + 19.82^2} \approx 30.3\ m/s(slightly faster than launch — landed below launch height)
Important notes
1. Air drag ignored. For light or fast objects in the real world, this matters.
2. Symmetric trajectory: if launch height = landing height, time up = time down, landing speed = launch speed.
3. 45° maximum range — only if launch height = landing height. From higher ground, the optimal angle is less than 45°.
4. With air resistance, 45° is not optimal — usually less (~35-40°).
What you should be able to do
- Decompose initial velocity into
x,ycomponents - Compute peak time, peak height, flight time, range
- Use the parabolic trajectory equation
- Optimize the launch angle for maximum range
Preview of §4.4
The second classical 2D motion: uniform circular motion. Constant speed but continuously changing direction — hence centripetal acceleration. §4.4.
📚 See also: Halliday Vol 1, Ch 4, §4.3 — Projectile Motion.
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