1. Write the standard projectile formulas

For a projectile launched with speed $u$ at angle $\theta$:

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

$H = \frac{u^2 \sin^2 \theta}{2g}$

where $R$ is the horizontal range and $H$ is the maximum height.

2. Use the given condition

The question says that the horizontal range is three times the maximum height:

$R = 3H$

Substitute the formulas:

$\frac{u^2 \sin 2\theta}{g} = 3 \cdot \frac{u^2 \sin^2 \theta}{2g}$

Cancel $\frac{u^2}{g}$ from both sides:

$\sin 2\theta = \frac{3}{2}\sin^2 \theta$

Now use

$\sin 2\theta = 2\sin\theta\cos\theta$

So,

$2\sin\theta\cos\theta = \frac{3}{2}\sin^2\theta$

Multiply by $2$:

$4\sin\theta\cos\theta = 3\sin^2\theta$

For non-zero projection, $\sin\theta \neq 0$, so divide by $\sin\theta$:

$4\cos\theta = 3\sin\theta$

Thus,

$\tan\theta = \frac{4}{3}$

3. Find $\sin 2\theta$

If $\tan\theta = \frac{4}{3}$, we can take a right triangle with sides:

• opposite $= 4$
• adjacent $= 3$
• hypotenuse $= 5$

Hence,

$\sin\theta = \frac{4}{5}, \qquad \cos\theta = \frac{3}{5}$

Therefore,

$\sin 2\theta = 2\sin\theta\cos\theta = 2\cdot \frac{4}{5} \cdot \frac{3}{5} = \frac{24}{25}$

4. Compute the range

$R = \frac{u^2 \sin 2\theta}{g} = \frac{u^2}{g} \cdot \frac{24}{25} = \frac{24u^2}{25g}$

Comparing with

$R = \frac{n u^2}{25g}$

we get

$n = 24$

5. Check with options

• A: $6$ ❌
• B: $12$ ❌
• C: $18$ ❌
• D: $24$ ✅

So the correct option is D.