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Part of ME-02 — Kinematics

Reasoning Chain — Why Complementary Angles Give Same Range

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Question: Why do θ = 30° and θ = 60° give the same range for the same launch speed u?

Chain:

R=u2sin⁡2θgR = \frac{u^2\sin 2\theta}{g}

→ For θ = 30°: R30=u2sin⁡(60°)g=u2×0.866gR_{30} = \frac{u^2\sin(60°)}{g} = \frac{u^2 \times 0.866}{g}

→ For θ = 60°: R60=u2sin⁡(120°)g=u2×0.866gR_{60} = \frac{u^2\sin(120°)}{g} = \frac{u^2 \times 0.866}{g}

→ sin⁡(2×30°)=sin⁡60°\sin(2 \times 30°) = \sin 60° and sin⁡(2×60°)=sin⁡120°\sin(2 \times 60°) = \sin 120°

→ But sin⁡60°=sin⁡(180°−60°)=sin⁡120°\sin 60° = \sin(180° - 60°) = \sin 120° (supplementary angles have equal sines)

→ Therefore R30=R60R_{30} = R_{60} ✓

Why heights differ:

H=u2sin⁡2θ2gH = \frac{u^2\sin^2\theta}{2g}

→ H30=u2sin⁡230°2g=u2(0.25)2gH_{30} = \frac{u^2\sin^2 30°}{2g} = \frac{u^2(0.25)}{2g}

→ H60=u2sin⁡260°2g=u2(0.75)2gH_{60} = \frac{u^2\sin^2 60°}{2g} = \frac{u^2(0.75)}{2g}

→ H60H30=0.750.25=3\frac{H_{60}}{H_{30}} = \frac{0.75}{0.25} = 3

→ The steeper angle (60°) converts more initial speed into vertical motion → rises 3× higher

Takeaway: Equal ranges arise from equal sin⁡(2θ)\sin(2\theta) values (supplementary 2θ angles). But sin⁡2θ\sin^2\theta values differ — hence unequal heights. This is a pure trigonometric result, not a coincidence.

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