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Projectile Motion

Source lecture(s): SC133 Lec 4

Intuition

Throw a ball: gravity pulls straight down and nothing pulls sideways. So the horizontal motion coasts at constant velocity while the vertical motion is ordinary free fall — two independent 1-D problems sharing a clock. The graceful parabola is just a straight line (horizontal) plus a dropped ball (vertical), superimposed. Galileo's famous demonstration: a bullet fired horizontally and one dropped at the same instant hit the ground together.

The equations

Launch speed \(v_0\) at angle \(\theta\) above the horizontal (no air resistance):

\[x(t) = v_0\cos\theta\; t \qquad\qquad y(t) = v_0\sin\theta\; t - \tfrac12 g t^2\]
\[v_x = v_0\cos\theta \;(\text{constant}) \qquad v_y = v_0\sin\theta - g t\]

Eliminating \(t\) gives the trajectory — a parabola:

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

The classic results (level ground)

Quantity Formula Maximized when
Time of flight \(T = \dfrac{2v_0\sin\theta}{g}\) \(\theta = 90°\)
Maximum height \(H = \dfrac{v_0^2\sin^2\theta}{2g}\) \(\theta = 90°\)
Range \(R = \dfrac{v_0^2\sin 2\theta}{g}\) \(\theta = 45°\)

Complementary angles (\(30°\) and \(60°\)) give the same range — one flies high and slow, the other low and fast.

The parabola of safety

Fire projectiles at fixed speed \(v_0\) in every direction. No shot can ever cross the envelope

\[y = \frac{v_0^2}{2g} - \frac{g\,x^2}{2 v_0^2}\]

— the parabola of safety. Anything outside it is unreachable. (This envelope is the star of the PHY653 warm-up simulation.)

Worked example: clearing a wall

A ball is kicked at \(20\,\text{m/s}\), \(45°\), toward a \(5\,\text{m}\) wall \(30\,\text{m}\) away. Does it clear?

At \(x = 30\): \(t = 30/(20\cos45°) = 2.12\,\text{s}\). Height: \(y = 20\sin45°(2.12) - 4.9(2.12)^2 = 30.0 - 22.0 = 8.0\,\text{m}\) — clears by 3 m. ✓

Try it live

The projectile playground lets you drag the launch angle and speed, watch range and height respond, and see the parabola of safety emerge from many trajectories.

Common mistakes

  • Letting gravity touch the horizontal motion. \(v_x\) never changes (without drag); only \(v_y\) does.
  • Using \(45°\) as optimal when launch and landing heights differ — from a cliff the optimum is less than 45°.
  • Forgetting the two motions share time \(t\). The link between \(x\) and \(y\) equations is the common clock — solve for \(t\) in one, substitute into the other.
  • Real balls feel air drag: trajectories are shorter and asymmetric, and the optimum angle drops to ~35° for a golf ball.

Knowledge graph position

Prerequisites: Kinematics, Vectors. Leads to: Circular motion, orbital motion — a cannonball fired fast enough is a satellite.

Quiz

Q1 (computational). A projectile launched at \(\theta = 30°\), \(v_0 = 40\,\text{m/s}\) on level ground: find the range.

Answer

\(R = v_0^2\sin(2\theta)/g = 1600\times\sin 60°/9.8 = 1600(0.866)/9.8 \approx 141\,\text{m}\).

Q2 (conceptual). At the top of its arc, is the projectile's speed zero?

Answer

No — only \(v_y = 0\). It still moves horizontally at \(v_0\cos\theta\); speed is at its minimum, not zero. (Contrast a ball thrown straight up.)

Q3 (conceptual). Two identical balls: one dropped, one launched horizontally from the same height. Which lands first?

Answer

They tie. Vertical motion is identical (same initial \(v_y = 0\), same \(g\)); horizontal velocity is irrelevant to the fall — the independence of components.

Q4 (multiple choice). Doubling the launch speed multiplies the level-ground range by: (a) 2 (b) \(\sqrt2\) (c) 4

Answer

(c) — \(R \propto v_0^2\). This quadratic sensitivity is why muzzle velocity dominates artillery design.