Two projectiles launched at complementary angles — prove same range

easy • 3 min read
Tags Projectile Motion

Question

Two projectiles are launched with the same speed but at angles θ\theta and (90°−θ)(90° - \theta). Prove that they have the same horizontal range.

Solution — Step by Step

The horizontal range of a projectile launched with initial speed uu at angle θ\theta (assuming level ground and ignoring air resistance) is:

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

For the first projectile at angle θ\theta:

R1=u2sin⁡2θgR_1 = \frac{u^2 \sin 2\theta}{g}

For the second projectile at angle (90°−θ)(90° - \theta):

R2=u2sin⁡[2(90°−θ)]g=u2sin⁡(180°−2θ)gR_2 = \frac{u^2 \sin[2(90° - \theta)]}{g} = \frac{u^2 \sin(180° - 2\theta)}{g}

Use the identity: sin⁡(180°−x)=sin⁡x\sin(180° - x) = \sin x

Therefore:

sin⁡(180°−2θ)=sin⁡2θ\sin(180° - 2\theta) = \sin 2\theta R2=u2sin⁡2θg=R1R_2 = \frac{u^2 \sin 2\theta}{g} = R_1

Therefore R1=R2R_1 = R_2. Proved.

Both projectiles have the same range when launched at complementary angles with the same initial speed.

Why This Works

The range formula R=u2sin⁡2θ/gR = u^2\sin 2\theta/g depends on sin⁡2θ\sin 2\theta. Since sin⁡(180°−x)=sin⁡x\sin(180° - x) = \sin x, we have sin⁡(2θ)=sin⁡(180°−2θ)=sin⁡(2(90°−θ))\sin(2\theta) = \sin(180° - 2\theta) = \sin(2(90° - \theta)). Complementary angles θ\theta and 90°−θ90° - \theta produce the same value of sin⁡2θ\sin 2\theta, hence the same range.

Geometrically: one angle launches the projectile “flatter” (more horizontal), the other launches it “steeper” (more vertical). The steeper angle takes longer in the air but covers less horizontal distance per unit time. These two effects exactly cancel — nature is elegantly balanced.

Example pairs:

  • 30° and 60° → sin⁡60°=sin⁡120°=3/2\sin 60° = \sin 120° = \sqrt{3}/2 → same range
  • 25° and 65° → sin⁡50°=sin⁡130°\sin 50° = \sin 130° → same range
  • 45° and 45° → maximum range (this is the only case where both angles are equal)

This result is a direct consequence of the sine rule: sin⁡x=sin⁡(180°−x)\sin x = \sin(180° - x). In JEE MCQs, if you see two launch angles that add up to 90°, you immediately know their ranges are equal — no calculation needed. The question might give you different time of flight (which WILL differ: T=2usin⁡θ/gT = 2u\sin\theta/g changes with θ\theta) and ask if the ranges are equal — they are.

Extension — Different Maximum Heights

While ranges are equal, maximum heights differ:

For θ\theta: H1=u2sin⁡2θ2gH_1 = \frac{u^2\sin^2\theta}{2g}

For (90°−θ)(90° - \theta): H2=u2sin⁡2(90°−θ)2g=u2cos⁡2θ2gH_2 = \frac{u^2\sin^2(90°-\theta)}{2g} = \frac{u^2\cos^2\theta}{2g}

H1+H2=u2(sin⁡2θ+cos⁡2θ)2g=u22g=Rmax2H_1 + H_2 = \frac{u^2(\sin^2\theta + \cos^2\theta)}{2g} = \frac{u^2}{2g} = \frac{R_{max}}{2}

(Since Rmax=u2/gR_{max} = u^2/g at 45°)

Also: H1⋅H2=u4sin⁡2θcos⁡2θ4g2=u4sin⁡22θ16g2=R216H_1 \cdot H_2 = \frac{u^4\sin^2\theta\cos^2\theta}{4g^2} = \frac{u^4\sin^2 2\theta}{16g^2} = \frac{R^2}{16}

So R=4H1H2R = 4\sqrt{H_1 H_2} — a useful result relating range to heights at complementary angles.

Common Mistake

Students sometimes write sin⁡(90°−θ)=cos⁡θ\sin(90° - \theta) = \cos\theta and then apply this to get R2=u2sin⁡(90°−θ)/g=u2cos⁡θ/gR_2 = u^2\sin(90°-\theta)/g = u^2\cos\theta/g, confusing sin⁡(2(90°−θ))\sin(2(90°-\theta)) with sin⁡(90°−θ)\sin(90°-\theta). The angle in the range formula is 2θ2\theta, not θ\theta. So for angle (90°−θ)(90°-\theta), we need sin⁡(2(90°−θ))=sin⁡(180°−2θ)=sin⁡2θ\sin(2(90°-\theta)) = \sin(180°-2\theta) = \sin 2\theta — NOT sin⁡(90°−θ)=cos⁡θ\sin(90°-\theta) = \cos\theta.

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