The horizontal range of a projectile fired at an angle of $15^\circ$ is 50 m. If it is fired with the same speed at an angle of $45^\circ$, its range will be (a) 60 m (b) 71 m (c) 100 m (d) 141 m

The horizontal range of a projectile fired at an angle of 15^° is 50 m. If it is fired with the same speed at

NCERT Exemplar · Q5

15^\circ

(a) 60 m

(b) 71 m

(d) 141 m

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Concept understanding — Projectile Range Symmetry

Projectile Range Symmetry

Imagine you're standing in a field and you throw a ball as hard as you can. You want it to land as far away as possible. Intuitively, you'd probably throw it at a 45° angle — and you'd be right. But here's the surprising part: if you throw it at 30° or at 60°, the ball lands at exactly the same distance.

That's the core idea of range symmetry.

The Intuition

Think about what happens when you launch a projectile at a shallow angle (say 20°). It has a large horizontal component of velocity, so it moves fast sideways — but it doesn't stay in the air very long because it barely goes upward. The range is limited by the short flight time.

Now think about a steep angle (say 70°). The ball goes high up, so it stays in the air a long time — but its horizontal speed is small because most of the launch velocity is directed upward. Again, the range is limited, this time by the low horizontal speed.

At 45°, you get the best trade-off: decent horizontal speed and decent flight time. That gives the maximum range.

But notice something: 20° and 70° are complementary angles — they add up to 90°. And they give the same range. So do 30° and 60°, 10° and 80°, and so on. The only exception is 45°, which is its own complement (45° + 45° = 90°), and it gives the maximum.

The Precise Statement

R(\theta) = \frac{u^2 \sin 2\theta}{g}

where $u$ is the launch speed, $\theta$ is the launch angle measured from the horizontal, and $g$ is the acceleration due to gravity.

Range symmetry says: for any launch angle $\theta$ (between 0° and 90°), the range at angle $\theta$ equals the range at angle $90° - \theta$ .

R(\theta) = R(90° - \theta)

Why It Works

Look at the formula. The range depends on $\sin 2\theta$ . Now:

\sin[2(90° - \theta)] = \sin(180° - 2\theta) = \sin 2\theta

Since $\sin(180° - x) = \sin x$ for any angle $x$ , the two ranges are identical. The sine function is symmetric about 90°, and that symmetry passes directly to the range.

Note

This symmetry holds only when launch and landing are at the same height. If you're throwing from a cliff or onto a slope, the symmetry breaks — the formula changes.

A Quick Example

A cricketer throws a ball at 20 m/s. At 30°, the range is:

R = \frac{(20)^2 \sin 60°}{9.8} = \frac{400 \times 0.866}{9.8} \approx 35.3 \text{ m}

At 60° (the complement), the range is:

R = \frac{400 \times \sin 120°}{9.8} = \frac{400 \times 0.866}{9.8} \approx 35.3 \text{ m}

Same number. At 45°, you get:

R = \frac{400 \times \sin 90°}{9.8} = \frac{400 \times 1}{9.8} \approx 40.8 \text{ m}

That's the maximum.

Common Mistake to Avoid …

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The range of a projectile depends on $\sin 2\theta$ . For the same speed, $R \propto \sin 2\theta$ . At $15^\circ$ , $\sin 30^\circ = 0.5$ gives $R = 50$ m; at $45^\circ$ , $\sin 90^\circ = 1$ doubles the range to $100$ m.

The key insight here is range symmetry — but not the symmetry you might first think of. Many students remember that $15^\circ$ and $75^\circ$ give the same range (since $\sin 30^\circ = \sin 150^\circ$ ). That’s true, but it’s not what this problem uses. Instead, we compare two angles with the same launch speed, and the range formula tells us everything.

The horizontal range of a projectile launched with speed $u$ at an angle $\theta$ is:

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

Since $u$ and $g$ are fixed in both cases, the range is directly proportional to $\sin 2\theta$ . That’s the only thing that changes.

For $\theta = 15^\circ$ , we have $2\theta = 30^\circ$ , so $\sin 30^\circ = \frac{1}{2}$ . The given range is $R_1 = 50$ m. Therefore:

50 = \frac{u^2}{g} \cdot \frac{1}{2} \quad \Rightarrow \quad \frac{u^2}{g} = 100

For $\theta = 45^\circ$ , we have $2\theta = 90^\circ$ , so $\sin 90^\circ = 1$ . The new range $R_2$ is:

R_2 = \frac{u^2}{g} \cdot 1 = 100 \text{ m}

That’s it — no need to compute $u$ or $g$ separately. The factor $\frac{u^2}{g}$ cancels neatly. …

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Q.The horizontal range of a projectile fired at an angle of $15^\circ$ is 50 m. If it is fired with the same speed at an angle of $45^\circ$ , its range will be

Concept understanding — Projectile Range Symmetry

Projectile Range Symmetry

The Intuition

The Precise Statement

Why It Works

A Quick Example

Common Mistake to Avoid …

Concept: Range Ratio at Fixed Speed, Without Ever Solving for $u^2/g$

Steps

Q.The horizontal range of a projectile fired at an angle of 15∘15^\circ15∘ is 50 m. If it is fired with the same speed at an angle of 45∘45^\circ45∘, its range will be

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Q.The horizontal range of a projectile fired at an angle of $15^\circ$ is 50 m. If it is fired with the same speed at an angle of $45^\circ$ , its range will be

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