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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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
$\tan^{-1}(\sqrt{7})$
›Reveal solutionSolution
The key idea is to use energy conservation to relate speed at half the maximum height to the launch speed. The value of $n$ is $\frac{3}{4}$ , so the correct option is (D).

We are told a body is projected from the ground at an angle $\theta = \tan^{-1}(\sqrt{7})$ . That means $\tan\theta = \sqrt{7}$ , so we can find $\sin\theta$ and $\cos\theta$ :
$\sin\theta = \frac{\sqrt{7}}{\sqrt{1+7}} = \frac{\sqrt{7}}{\sqrt{8}} = \frac{\sqrt{7}}{2\sqrt{2}}, \quad \cos\theta = \frac{1}{\sqrt{8}} = \frac{1}{2\sqrt{2}}.$
The launch speed is $u$ . At half the maximum height, the speed is $v = n u$ . We need $n$ .

Concept & Intuition:

In projectile motion, horizontal velocity is constant (no horizontal force). Vertical motion is governed by gravity, so vertical speed changes with height. Energy conservation is the cleanest way: the loss in kinetic energy equals the gain in gravitational potential energy. At half the maximum height, the potential energy is half of what it would be at the top, so the kinetic energy is reduced by that amount. This directly gives the speed ratio without solving for time.

Step-by-step solution:

Maximum height reached: The vertical component of initial velocity is $u_y = u \sin\theta$ . At the top, vertical speed is zero. Using $v_y^2 = u_y^2 - 2gH$ :

$0 = u^2 \sin^2\theta - 2gH \quad\Rightarrow\quad H = \frac{u^2 \sin^2\theta}{2g}.$

Half of maximum height:

Let $h = \frac{H}{2} = \frac{u^2 \sin^2\theta}{4g}$ .

Energy conservation between launch and height $h$ :

At launch: total mechanical energy = $\frac{1}{2} m u^2$ (taking ground as zero potential).

At height $h$ : kinetic energy = $\frac{1}{2} m v^2$ , potential energy = $m g h$ .

So:

$\frac{1}{2} m u^2 = \frac{1}{2} m v^2 + m g h.$
Cancel $m$ and multiply by 2:
$u^2 = v^2 + 2g h.$

Substitute $h$ :
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TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
$= 10\, \mathrm{ms}^{-2}$
›Reveal solutionSolution
The key is to first find the launch speed from the given maximum range on level ground, then treat the tower throw as a horizontal projectile with that same speed; the horizontal distance is speed times the time to fall 25 m, giving 40 m.

Concept and Intuition

The problem gives two scenarios with the same initial speed.

On level ground, maximum range occurs at a $45^\circ$ launch angle.

From a tower, the ball is thrown horizontally — so its entire initial velocity is horizontal, and it falls under gravity. The horizontal distance covered is simply the horizontal speed multiplied by the time it takes to hit the ground. We can extract the speed from the first scenario and then apply it to the second.

Step-by-step solution

Find the initial speed from the maximum range on level ground. For a projectile launched from ground level, the range is

$R = \frac{u^2 \sin 2\theta}{g}.$
Maximum range occurs when $\sin 2\theta = 1$ (i.e., $\theta = 45^\circ$ ), giving
$R_{\text{max}} = \frac{u^2}{g}.$
Here $R_{\text{max}} = 32\ \text{m}$ and $g = 10\ \text{m/s}^2$ . So
$\frac{u^2}{10} = 32 \quad\Rightarrow\quad u^2 = 320 \quad\Rightarrow\quad u = \sqrt{320} = 8\sqrt{5}\ \text{m/s}.$

Now consider the horizontal throw from the tower. The ball is thrown horizontally with the same speed $u = 8\sqrt{5}\ \text{m/s}$ . The tower height is $h = 25\ \text{m}$ . The time to fall vertically from rest is found from

$h = \frac{1}{2} g t^2 \quad\Rightarrow\quad 25 = \frac{1}{2} \cdot 10 \cdot t^2 = 5 t^2.$
Hence
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TG EAPCET 2023Set eng-2023-05-14-FN1 markMCQ
$Q. A player can throw a ball to a maximum horizontal distance of 80 m. If he throws the ball vertically with the same velocity, then the maximum height reached by the ball is (A) 160 m (B) 60 m (C) 20 m (D) 40 m$
›Reveal solutionSolution
The key idea is that the same initial speed gives a maximum range of $80\,\text{m}$ when thrown at $45^\circ$ , and that speed determines the maximum height in vertical throw as $40\,\text{m}$ . The correct option is (D).

The problem connects two classic projectile scenarios: maximum horizontal range and vertical throw. The crucial insight is that the same initial speed is used in both cases. For a given launch speed $u$ , the maximum range on level ground occurs at a $45^\circ$ launch angle, and the maximum height when thrown straight up is simply $u^2/(2g)$ . So if we can find $u^2/(2g)$ from the range data, we have the answer directly.

Relate maximum range to initial speed. For a projectile launched at angle $\theta$ with speed $u$ , the range is

$R = \frac{u^2 \sin 2\theta}{g}.$
The maximum range occurs when $\sin 2\theta = 1$ , i.e., $\theta = 45^\circ$ . Thus
$R_{\text{max}} = \frac{u^2}{g}.$
We are told $R_{\text{max}} = 80\,\text{m}$ , so
$\frac{u^2}{g} = 80 \quad \Rightarrow \quad u^2 = 80g.$

Find maximum height for vertical throw. When the ball is thrown straight upward with the same speed $u$ , the maximum height $H$ is given by

$H = \frac{u^2}{2g}.$
Substitute $u^2 = 80g$ :
$H = \frac{80g}{2g} = 40\,\text{m}.$

Interpret the result. …
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TG EAPCET 2022Set eng-2022-07-18-AN1 markMCQ
$(3\hat{i} + 4\hat{j})\, m/s$
›Reveal solutionSolution
The trajectory of a projectile is determined by its initial velocity components and gravity. By deriving the standard equation of trajectory and comparing it with the given form, we find that the value of $\gamma$ is $\boxed{-5}$ .

The path followed by a projectile, known as its trajectory, is a parabola. This parabolic path arises from the combination of uniform horizontal motion (constant velocity) and uniformly accelerated vertical motion (due to gravity). To find the equation of this trajectory, we express the horizontal and vertical positions of the projectile as functions of time and then eliminate time.

Here's how we can determine the value of $\gamma$ :

Identify initial velocity components:

The initial velocity of the projectile is given as $\vec{u} = (3\hat{i} + 4\hat{j})\, m/s$ .

This means the initial horizontal velocity component is $u_x = 3\, m/s$ , and the initial vertical velocity component is $u_y = 4\, m/s$ .

The acceleration due to gravity is $g = 10\, m/s^2$ , acting downwards.

Formulate equations of motion:

For horizontal motion, there is no acceleration (assuming air resistance is negligible).

The horizontal displacement $x$ after time $t$ is given by:

$x = u_x t$

$x = 3t \quad \implies \quad t = \frac{x}{3} \quad \text{(Equation 1)}$

For vertical motion, the acceleration is $-g$ (downwards).

The vertical displacement $y$ after time $t$ is given by:

$y = u_y t - \frac{1}{2} g t^2$

$y = 4t - \frac{1}{2} (10) t^2$

$y = 4t - 5t^2 \quad \text{(Equation 2)}$

Derive the equation of trajectory:

Substitute Equation 1 into Equation 2 to eliminate $t$ :

$y = 4\left(\frac{x}{3}\right) - 5\left(\frac{x}{3}\right)^2$

$y = \frac{4x}{3} - 5\left(\frac{x^2}{9}\right)$

$y = \frac{4x}{3} - \frac{5x^2}{9}$
$u_x$
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TG EAPCET 2021Set eng-2021-08-06-AN1 markMCQ
$h$
›Reveal solutionSolution
The range of a TV tower is proportional to the square root of its height. Increasing height from $h$ to $\frac{3}{2}h$ multiplies the range by $\sqrt{\frac{3}{2}}$ , so the change in range is $\left(\sqrt{\frac{3}{2}} - 1\right)d$ .

The key concept here is the line-of-sight propagation of TV signals. A transmitting tower of height $h$ can only be seen by a receiver up to the point where the Earth’s curvature blocks the line of sight. For a spherical Earth of radius $R$ , the maximum distance $d$ (the range) from the tower to the horizon is given by the geometry of a right triangle: the tower height is one leg, the Earth’s radius is another, and the line of sight is the hypotenuse.

From simple geometry, using the approximation $h \ll R$ , we get the classic result:
$d = \sqrt{2 R h}$
This tells us that $d \propto \sqrt{h}$ . So if the height changes, the range changes as the square root of the height ratio.

Now let’s work through the problem step by step.

Write the initial range. For height $h$ , the range is

$d = \sqrt{2 R h}$

Write the new range for the increased height. New height $h' = \frac{3}{2}h$ . So the new range $d'$ is

$d' = \sqrt{2 R \cdot \frac{3}{2}h} = \sqrt{\frac{3}{2} \cdot 2 R h} = \sqrt{\frac{3}{2}} \cdot \sqrt{2 R h}$

Express $d'$ in terms of the original $d$ . Since $d = \sqrt{2 R h}$ , we have

$d' = \sqrt{\frac{3}{2}} \, d$

Find the change in range. The change is $d' - d$ . Substituting: …
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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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