Projectiles: Hitting a Target

The question

You are a trebuchet operator. The enemy gate sits 50 m away on level ground and your machine always launches stones at exactly 25 m/s. You cannot change the speed — only the angle. Is there a single 'correct' angle that drops the stone on the gate, or could two completely different angles both score a direct hit?

There are TWO. A low, fast, flat shot and a high, lazy, arcing shot land at the very same spot — they are mirror images about 45°. Master this and you can pick the angle that clears a castle wall, sneak under a low ceiling, or reach the farthest target your machine can possibly hit.

The key idea

On LEVEL ground (the projectile lands at the same height it was launched from), the horizontal range is fixed by launch speed and angle alone: R = v₀²sin(2θ)/g. To hit a target a known distance D away you invert this to find the angle, and because sin is symmetric you get TWO solutions that add to 90°. The range is greatest at 45°, and the impact speed and angle mirror the launch.

The range formula is just the two motions stitched together. Flight time on level ground is $t = 2v_0\sin\theta/g$ (up–down symmetry holds because it lands at launch height), and horizontal distance is $R = v_0\cos\theta\cdot t = \dfrac{2v_0^2\sin\theta\cos\theta}{g}$. The identity $2\sin\theta\cos\theta = \sin 2\theta$ collapses this to $R = \dfrac{v_0^2\sin 2\theta}{g}$. **The two-angle solution:** to hit distance $D$ you need $\sin 2\theta = Dg/v_0^2$. Since $\sin$ gives the same value at an angle and its supplement, both $2\theta$ and $180^\circ-2\theta$ work, so $\theta$ AND $90^\circ-\theta$ both hit the target — a low flat shot and a high lobbed shot. **Why 45° is the champion:** $\sin 2\theta$ can be at most 1, reached when $2\theta = 90^\circ$, so the farthest any launch speed can reach is $R_\text{max} = v_0^2/g$ at exactly $45^\circ$. **Impact:** on level ground the projectile comes back to its starting height, so energy conservation forces the landing speed to equal $v_0$, and the velocity makes the same angle below horizontal as it was launched above. **The fine print:** this ALL assumes level ground. Launch from or onto a different height and the symmetry breaks — you must go back to the kinematic equations and the simple range formula no longer applies.

The formula

• ·R = v₀²sin(2θ)/g gives the level-ground range
• ·where v₀ = launch speed (m/s)
• ·θ = launch angle
• ·g = 9.81 m/s². To hit a target at distance D
• ·set R = D and solve sin(2θ) = Dg/v₀². Taking arcsin gives 2θ
• ·so θ = ½·arcsin(Dg/v₀²); the SECOND valid angle is 90°−θ
• ·because sin(180°−2θ) equals sin(2θ). The range is largest when sin(2θ) = 1
• ·i.e. θ = 45°
• ·giving R_max = v₀²/g. On level ground the stone returns to launch height
• ·so by energy its impact SPEED equals the launch speed v₀
• ·and by symmetry it strikes the ground at angle θ below the horizontal (the mirror of the launch).

What to remember

• ·On level ground the range is R = v₀²sin(2θ)/g; to hit a target at distance D solve sin(2θ) = Dg/v₀², which yields TWO complementary angles θ and 90°−θ (a flat shot and a lobbed shot).
• ·Maximum range R_max = v₀²/g occurs at exactly 45°; a target beyond that distance is unreachable at that speed, and a target exactly at R_max is hit by only one angle (45°).
• ·On level ground the projectile returns to launch height, so its impact speed equals the launch speed v₀ and it strikes at the same angle below the horizontal as it was launched above.