Cannon Ball Trajectory

A cannon at ground level fires a ball at speed $v_0$ and angle $\theta$ above the horizontal. Where does the cannonball land? This is the classical artillery problem — and it was the first projectile question solved with calculus, by Galileo around 1638.

The Physics

For ground-to-ground projectile motion, the three big quantities follow directly from decomposing the launch velocity into components:

$v_{0x} = v_0\cos\theta \qquad v_{0y} = v_0\sin\theta$

$T = \dfrac{2 v_0\sin\theta}{g} \qquad H = \dfrac{v_0^{2}\sin^{2}\theta}{2g} \qquad R = \dfrac{v_0^{2}\sin(2\theta)}{g}$

Step-by-Step Solution

Given:

• Muzzle velocity: $v_0 = 100\;\text{m/s}$
• Cannon angle: $\theta = 30°$
• Gravity: $g = 9.81\;\text{m/s}^{2}$

Find: Range, peak altitude, and total flight time.

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Step 1 — Decompose the muzzle velocity into components.

$v_{0x} = 100\cos 30° = 100 \times \dfrac{\sqrt{3}}{2} \approx 86.60\;\text{m/s}$

$v_{0y} = 100\sin 30° = 100 \times 0.5 = 50.00\;\text{m/s}$

Step 2 — Find the time of flight.

Using $T = 2 v_{0y}/g$:

$T = \dfrac{2 \times 50.00}{9.81} = \dfrac{100}{9.81} \approx 10.194\;\text{s}$

Step 3 — Find the maximum altitude.

The cannonball reaches its peak halfway through the flight, at $t_{\text{up}} = T/2 \approx 5.097\;\text{s}$. Using $H = v_{0y}^{2}/(2g)$:

$H = \dfrac{(50.00)^{2}}{2 \times 9.81} = \dfrac{2500}{19.62} \approx 127.42\;\text{m}$

Step 4 — Find the horizontal range.

The horizontal velocity is constant, so:

$R = v_{0x} \cdot T = 86.60 \times 10.194 \approx 882.83\;\text{m}$

We can also verify with the closed form:

$R = \dfrac{v_0^{2}\sin(2\theta)}{g} = \dfrac{(100)^{2}\sin 60°}{9.81} = \dfrac{10000 \times 0.8660}{9.81} \approx 882.83\;\text{m} \checkmark$

Step 5 — Find the impact velocity.

Horizontal: $v_x = 86.60\;\text{m/s}$ (unchanged).

Vertical: $v_y = -g\,T/2 \times 2 / 2 = -50\;\text{m/s}$ (the negative of $v_{0y}$, by symmetry).

Impact speed: $|\vec v| = \sqrt{(86.60)^{2} + (50)^{2}} = \sqrt{7500 + 2500} = \sqrt{10000} = 100\;\text{m/s}$.

(Energy conservation: launch and landing at the same elevation means the same speed.)

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Answer: The cannonball travels a horizontal range of $\approx 882.83\;\text{m}$ (almost a kilometer), reaches a peak altitude of $\approx 127.42\;\text{m}$ (about a 40-story building), and stays in the air for $\approx 10.19\;\text{s}$. It strikes the ground at the same speed it left — 100 m/s.

The Historical Context

Galileo was the first to realize, around 1638, that projectile motion was a parabola. Before that, cannoneers used trial-and-error firing tables. After Galileo, you could predict the impact point if you knew the launch speed and angle — a transformative military advantage.

Try It

• Drop the angle to a flat 5° — short range, fast impact, very low arc.
• Push it to 85° — the ball climbs nearly straight up (peak ≈ 506 m!) and lands very close to the cannon.
• Drag the velocity to 200 m/s and watch the range explode to ~3.5 km (since $R \propto v_0^{2}$).