A projectile is thrown with an initial velocity of `(a hati + hatj) ms^(-1)`. If the range of the projectile is twice the maximum height reached by it, then

Options

A. `a = 2b`
B. `b = a`
C. `b = 2a`
D. `b = 4a`

Explanation

To solve the problem step by step, we will analyze the motion of the projectile thrown with an initial velocity of \( \mathbf{u} = a \hat{i} + b \hat{j} \) m/s. We need to find the relationship between \( a \) and \( b \) given that the range \( R \) is twice the maximum height \( H_{\text{max}} \).

### Step 1: Identify the components of the initial velocity
The initial velocity \( \mathbf{u} \) can be broken down into its horizontal and vertical components:
- Horizontal component: \( u_x = a \)
- Vertical component: \( u_y = b \)

### Step 2: Calculate the maximum height \( H_{\text{max}} \)
The maximum height reached by the projectile can be calculated using the formula:
\[ H_{\text{max}} = \frac{u_y^2}{2g} \]
Substituting \( u_y = b \):
\[ H_{\text{max}} = \frac{b^2}{2g} \]

### Step 3: Calculate the time of flight \( T \)
The time of flight \( T \) for a projectile is given by:
\[ T = \frac{2u_y}{g} \]
Substituting \( u_y = b \):
\[ T = \frac{2b}{g} \]

### Step 4: Calculate the range \( R \)
The range of the projectile is given by:
\[ R = u_x \cdot T \]
Substituting \( u_x = a \) and \( T = \frac{2b}{g} \):
\[ R = a \cdot \frac{2b}{g} = \frac{2ab}{g} \]

### Step 5: Set up the relationship between range and maximum height
According to the problem, the range is twice the maximum height:
\[ R = 2H_{\text{max}} \]
Substituting the expressions for \( R \) and \( H_{\text{max}} \):
\[ \frac{2ab}{g} = 2 \cdot \frac{b^2}{2g} \]
This simplifies to:
\[ \frac{2ab}{g} = \frac{b^2}{g} \]

### Step 6: Simplify the equation
By multiplying both sides by \( g \) (assuming \( g \neq 0 \)):
\[ 2ab = b^2 \]
Assuming \( b \neq 0 \), we can divide both sides by \( b \):
\[ 2a = b \]

### Step 7: Final result
Thus, we find the relationship between \( a \) and \( b \):
\[ b = 2a \]

### Conclusion
The final answer is that \( b \) is equal to \( 2a \).

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