A particle is projected with a velocity v so that it's range on a horizontal plane is twice the greatest height attained if g is the acceleration due to gravity then find R

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.

Learn how to solve for the range of a projectile using the given initial velocity, angle of projection, and acceleration due to gravity. This problem involves applying trigonometric identities and the range formula of projectile motion. Suitable for high school students.

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