To solve for the range $R$ of a projectile given that the range is twice the greatest height attained, let's start by recalling the relevant equations of projectile motion. Let:

• $v$ be the initial velocity of the projectile,
• $\theta$ be the angle of projection,
• $g$ be the acceleration due to gravity.

The range $R$ of a projectile is given by:

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

The greatest height $H$ attained by the projectile is given by:

$H = \frac{v^2 \sin^2(\theta)}{2g}$

According to the problem, the range is twice the greatest height:

$R = 2H$

Substitute the expressions for $R$ and $H$:

$\frac{v^2 \sin(2\theta)}{g} = 2 \left( \frac{v^2 \sin^2(\theta)}{2g} \right)$

Simplify the equation:

$\frac{v^2 \sin(2\theta)}{g} = \frac{v^2 \sin^2(\theta)}{g}$

Cancel out the common terms ($v^2$ and $g$) on both sides:

$\sin(2\theta) = \sin^2(\theta)$

Using the trigonometric identity for $\sin(2\theta)$:

$2 \sin(\theta) \cos(\theta) = \sin^2(\theta)$

Divide both sides by $\sin(\theta)$ (assuming $\theta \neq 0$):

$2 \cos(\theta) = \sin(\theta)$

$\tan(\theta) = 2$

So, the angle of projection $\theta$ is:

$\theta = \tan^{-1}(2)$

Now, substitute $\theta = \tan^{-1}(2)$ back into the equation for the range $R$:

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

We know:

$\sin(2 \theta) = \sin(2 \tan^{-1}(2))$

Using the double angle identity for $\sin$:

$\sin(2 \theta) = \frac{2 \tan(\theta)}{1 + \tan^2(\theta)}$

Since $\tan(\theta) = 2$:

$\sin(2 \theta) = \frac{2 \cdot 2}{1 + 2^2} = \frac{4}{5}$

Therefore, the range $R$ is:

$R = \frac{v^2 \cdot \frac{4}{5}}{g} = \frac{4v^2}{5g}$

So, the range $R$ is:

$R = \frac{4v^2}{5g}$

Would you like more details or have any questions? Here are five more questions you can ask:

1. How do you find the maximum height of a projectile?
2. What are the kinematic equations used for projectile motion?
3. How does the angle of projection affect the range of a projectile?
4. Can you explain the derivation of the range formula in projectile motion?
5. What are some real-world applications of projectile motion?

Tip: In projectile motion problems, always break down the motion into horizontal and vertical components to simplify calculations.