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

Options

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

Answer

To solve the problem, we need to establish the relationship between the initial velocity components \( a \) and \( b \) given the condition that the range of the projectile is twice the maximum height reached.

### Step 1: Understand the initial velocity components
The initial velocity of the projectile is given as:
\[ \vec{u} = a \hat{i} + b \hat{j} \]
From this, we can identify:
- The horizontal component of velocity: \( u_x = a \)
- The vertical component of velocity: \( u_y = b \)

### Step 2: Relate the components to the initial speed
The initial speed \( u \) can be expressed as:
\[ u = \sqrt{a^2 + b^2} \]

### Step 3: Determine the range and maximum height formulas
The range \( R \) of a projectile is given by:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
And the maximum height \( H \) is given by:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]

### Step 4: Use the given condition
According to the problem, the range is twice the maximum height:
\[ R = 2H \]
Substituting the formulas for \( R \) and \( H \):
\[ \frac{u^2 \sin 2\theta}{g} = 2 \left(\frac{u^2 \sin^2 \theta}{2g}\right) \]
This simplifies to:
\[ \frac{u^2 \sin 2\theta}{g} = \frac{u^2 \sin^2 \theta}{g} \]
We can cancel \( \frac{u^2}{g} \) (assuming \( u \neq 0 \)):
\[ \sin 2\theta = \sin^2 \theta \]

### Step 5: Use the double angle identity
Using the identity \( \sin 2\theta = 2 \sin \theta \cos \theta \):
\[ 2 \sin \theta \cos \theta = \sin^2 \theta \]
Rearranging gives:
\[ \sin^2 \theta - 2 \sin \theta \cos \theta = 0 \]
Factoring out \( \sin \theta \):
\[ \sin \theta (\sin \theta - 2 \cos \theta) = 0 \]
This gives us two cases:
1. \( \sin \theta = 0 \) (not applicable for projectile motion)
2. \( \sin \theta = 2 \cos \theta \)

### Step 6: Find the relationship between \( a \) and \( b \)
From \( \sin \theta = 2 \cos \theta \), we can write:
\[ \tan \theta = \frac{\sin \theta}{\cos \theta} = 2 \]
Thus, we have:
\[ \frac{b}{a} = 2 \quad \text{(since } b = u \sin \theta \text{ and } a = u \cos \theta\text{)} \]
This implies:
\[ b = 2a \]

### Conclusion
The relationship between \( b \) and \( a \) is:
\[ b = 2a \]