Comparing Projectile Angles: 30°, 45°, 60°

Three balls are launched at the same speed but at three different angles: 30°, 45°, and 60°. Side-by-side they reveal one of the prettiest symmetries in projectile motion — and the famous result that complementary angles produce the same range.

The Physics

For launches at the same elevation:

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

Why complementary angles match: $\sin(2\theta) = \sin(180° - 2\theta) = \sin(2(90° - \theta))$. So $\theta$ and $90° - \theta$ give the same $\sin(2\theta)$ — and therefore the same range. The 45° launch sits exactly at the peak of $\sin(2\theta)$, so it goes the furthest.

Step-by-Step Solution

Given: $v_0 = 25\;\text{m/s}$, $g = 9.81\;\text{m/s}^{2}$, three angles: 30°, 45°, 60°.

Find: $R$, $H$, and $T$ for each launch.

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For $\theta = 30°$:

Velocity components:

$v_{0x} = 25\cos 30° = 25 \times 0.8660 \approx 21.65\;\text{m/s}$

$v_{0y} = 25\sin 30° = 25 \times 0.5000 = 12.50\;\text{m/s}$

Time of flight:

$T_{30} = \dfrac{2(12.50)}{9.81} = \dfrac{25}{9.81} \approx 2.548\;\text{s}$

Maximum height:

$H_{30} = \dfrac{(12.50)^{2}}{2(9.81)} = \dfrac{156.25}{19.62} \approx 7.96\;\text{m}$

Range:

$R_{30} = \dfrac{25^{2}\sin 60°}{9.81} = \dfrac{625 \times 0.8660}{9.81} \approx 55.16\;\text{m}$

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For $\theta = 45°$:

$v_{0x} = v_{0y} = 25\sin 45° \approx 17.68\;\text{m/s}$

$T_{45} = \dfrac{2(17.68)}{9.81} \approx 3.604\;\text{s}$

$H_{45} = \dfrac{(17.68)^{2}}{19.62} = \dfrac{312.5}{19.62} \approx 15.93\;\text{m}$

$R_{45} = \dfrac{625 \times \sin 90°}{9.81} = \dfrac{625}{9.81} \approx 63.71\;\text{m}$

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For $\theta = 60°$:

$v_{0x} = 25\cos 60° = 12.50\;\text{m/s}$

$v_{0y} = 25\sin 60° \approx 21.65\;\text{m/s}$

$T_{60} = \dfrac{2(21.65)}{9.81} \approx 4.414\;\text{s}$

$H_{60} = \dfrac{(21.65)^{2}}{19.62} = \dfrac{468.75}{19.62} \approx 23.89\;\text{m}$

$R_{60} = \dfrac{625 \times \sin 120°}{9.81} = \dfrac{625 \times 0.8660}{9.81} \approx 55.16\;\text{m}$

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Answer:

• 30°: $R \approx 55.16\;\text{m}$, $H \approx 7.96\;\text{m}$, $T \approx 2.55\;\text{s}$
• 45°: $R \approx 63.71\;\text{m}$, $H \approx 15.93\;\text{m}$, $T \approx 3.60\;\text{s}$ ← maximum range
• 60°: $R \approx 55.16\;\text{m}$, $H \approx 23.89\;\text{m}$, $T \approx 4.41\;\text{s}$

Notice that $R_{30} = R_{60}$ exactly (≈ 55.16 m) — they're complementary angles. The 60° ball reaches three times the height of the 30° ball and stays in the air 70% longer, but lands in the same spot.

Try It

• Watch the 45° (yellow) ball reach the farthest distance — it lands about 8.5 m beyond the other two.
• Notice the 30° (cyan) and 60° (pink) balls hit the ground at the same place — but the 60° ball arrives much later.
• Bump the velocity to see all three paths scale uniformly. Range scales with $v_0^{2}$, so doubling the speed quadruples the range.