A projectile is thrown with an initial velocity of `(xveci + y vecj) m//s` . If the range of the projectile is double the maximum height reached by it, then

Options

A. `x-2y`
B. `y=2x`
C. `x=y`
D. `y=4x`

Explanation

To solve the problem, we need to establish the relationship between the initial velocity components \( x \) and \( y \) given that the range \( R \) of the projectile is double the maximum height \( H \) reached by it.

### Step-by-Step Solution:

1. **Identify the Components of Initial Velocity:**
The initial velocity \( \vec{u} \) of the projectile can be expressed in terms of its components:
\[ \vec{u} = x \hat{i} + y \hat{j} \]
Here, \( x \) is the horizontal component and \( y \) is the vertical component of the initial velocity.

2. **Calculate the Range of the Projectile:**
The range \( R \) of a projectile launched at an angle \( \theta \) with initial velocity \( u \) is given by:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
We can express \( \sin 2\theta \) in terms of \( x \) and \( y \):
\[ \sin 2\theta = 2 \sin \theta \cos \theta \]
Where:
\[ \sin \theta = \frac{y}{u}, \quad \cos \theta = \frac{x}{u} \]
Therefore:
\[ R = \frac{u^2 \cdot 2 \left(\frac{y}{u}\right) \left(\frac{x}{u}\right)}{g} = \frac{2xy}{g} \]

3. **Calculate the Maximum Height of the Projectile:**
The maximum height \( H \) reached by the projectile is given by:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]
Substituting \( \sin \theta = \frac{y}{u} \):
\[ H = \frac{u^2 \left(\frac{y}{u}\right)^2}{2g} = \frac{y^2}{2g} \]

4. **Set Up the Relationship Between Range and Maximum Height:**
According to the problem, the range is double the maximum height:
\[ R = 2H \]
Substituting the expressions for \( R \) and \( H \):
\[ \frac{2xy}{g} = 2 \cdot \frac{y^2}{2g} \]
Simplifying this gives:
\[ 2xy = 2y^2 \]

5. **Simplify the Equation:**
Dividing both sides by 2 (assuming \( y \neq 0 \)):
\[ xy = y^2 \]
Rearranging gives:
\[ y^2 - xy = 0 \]
Factoring out \( y \):
\[ y(y - x) = 0 \]

6. **Find the Relationship Between \( x \) and \( y \):**
This gives us two solutions:
- \( y = 0 \) (not valid for projectile motion)
- \( y = x \)

Thus, the relationship between the components of the initial velocity is:
\[ y = x \]