The kinetic energy of a projectile at its highest position is `K`. If the range of the projectile is four times the height of the projectile, then the initial kinetic energy of the projectile is .

Options

A. `sqrt2K`
B. `2 K`
C. `4 K`
D. `2sqrt2 K`

Explanation

To solve the problem, we need to find the initial kinetic energy of a projectile given that its kinetic energy at the highest point is \( K \) and the range of the projectile is four times the height.

### Step-by-Step Solution:

1. **Understand the Given Information**:
- The kinetic energy at the highest point is \( K \).
- The range \( R \) is four times the height \( h \) of the projectile, i.e., \( R = 4h \).

2. **Write the Formulas for Range and Height**:
- The formula for the range \( R \) of a projectile is:
\[ R = \frac{2u^2 \sin \theta \cos \theta}{g} \]
- The formula for the maximum height \( h \) of a projectile is:
\[ h = \frac{u^2 \sin^2 \theta}{2g} \]

3. **Substitute the Relationship Between Range and Height**:
- Since \( R = 4h \), we can substitute the expression for \( h \) into the equation:
\[ \frac{2u^2 \sin \theta \cos \theta}{g} = 4 \left(\frac{u^2 \sin^2 \theta}{2g}\right) \]

4. **Simplify the Equation**:
- Cancel \( g \) and \( u^2 \) from both sides (assuming \( u \neq 0 \)):
\[ 2 \sin \theta \cos \theta = 4 \left(\frac{\sin^2 \theta}{2}\right) \]
- This simplifies to:
\[ 2 \sin \theta \cos \theta = 2 \sin^2 \theta \]
- Dividing both sides by 2:
\[ \sin \theta \cos \theta = \sin^2 \theta \]

5. **Rearranging the Equation**:
- Rearranging gives:
\[ \sin \theta \cos \theta - \sin^2 \theta = 0 \]
- Factoring out \( \sin \theta \):
\[ \sin \theta (\cos \theta - \sin \theta) = 0 \]
- This implies \( \sin \theta = 0 \) or \( \cos \theta = \sin \theta \). Since \( \sin \theta = 0 \) is not valid for projectile motion, we have:
\[ \cos \theta = \sin \theta \]
- This occurs when \( \theta = 45^\circ \).

6. **Calculate the Kinetic Energy at the Highest Point**:
- At the highest point, the vertical component of the velocity is zero, and the horizontal component remains:
\[ V = u \cos \theta \]
- Therefore, the kinetic energy at the highest point is:
\[ K = \frac{1}{2} m V^2 = \frac{1}{2} m (u \cos \theta)^2 \]
- Substituting \( \theta = 45^\circ \) (where \( \cos 45^\circ = \frac{1}{\sqrt{2}} \)):
\[ K = \frac{1}{2} m \left(u \cdot \frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} m \frac{u^2}{2} = \frac{1}{4} mu^2 \]

7. **Relate Initial Kinetic Energy to \( K \)**:
- The initial kinetic energy \( KE_{initial} \) is:
\[ KE_{initial} = \frac{1}{2} mu^2 \]
- From our expression for \( K \):
\[ K = \frac{1}{4} mu^2 \]
- Therefore, we can express \( \frac{1}{2} mu^2 \) in terms of \( K \):
\[ KE_{initial} = 2K \]

### Final Answer:
The initial kinetic energy of the projectile is \( 2K \).