If the maximum vertical height and horizontal ranges of a projectile are same, the angle of projection will be

Options

A. `45^(@)`
B. `tan^(-1) 4`
C. `tan^(-1) 2`
D. `30^(@)`

Answer

To solve the problem of finding the angle of projection when the maximum vertical height and horizontal range of a projectile are the same, we can follow these steps:

### Step 1: Understand the relationships
The maximum height \( H \) and the horizontal range \( R \) of a projectile launched at an angle \( \theta \) with an initial velocity \( u \) are given by the formulas:
- Maximum height:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]
- Horizontal range:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
where \( g \) is the acceleration due to gravity.

### Step 2: Set the equations equal
According to the problem, the maximum height is equal to the horizontal range:
\[ H = R \]
Substituting the formulas for \( H \) and \( R \):
\[ \frac{u^2 \sin^2 \theta}{2g} = \frac{u^2 \sin 2\theta}{g} \]

### Step 3: Simplify the equation
We can cancel \( u^2 \) and \( g \) from both sides (assuming \( u \neq 0 \) and \( g \neq 0 \)):
\[ \frac{\sin^2 \theta}{2} = \sin 2\theta \]

### Step 4: Use the double angle identity
Recall that \( \sin 2\theta = 2 \sin \theta \cos \theta \). Substituting this into the equation gives:
\[ \frac{\sin^2 \theta}{2} = 2 \sin \theta \cos \theta \]

### Step 5: Rearrange the equation
Multiply both sides by 2:
\[ \sin^2 \theta = 4 \sin \theta \cos \theta \]
Rearranging gives:
\[ \sin^2 \theta - 4 \sin \theta \cos \theta = 0 \]

### Step 6: Factor the equation
Factoring out \( \sin \theta \):
\[ \sin \theta (\sin \theta - 4 \cos \theta) = 0 \]
This gives us two cases:
1. \( \sin \theta = 0 \)
2. \( \sin \theta - 4 \cos \theta = 0 \)

### Step 7: Solve the first case
For \( \sin \theta = 0 \):
- This occurs at \( \theta = 0^\circ \) or \( 180^\circ \), which are not valid angles for projectile motion.

### Step 8: Solve the second case
For \( \sin \theta - 4 \cos \theta = 0 \):
\[ \sin \theta = 4 \cos \theta \]
Dividing both sides by \( \cos \theta \) gives:
\[ \tan \theta = 4 \]
Thus, we find:
\[ \theta = \tan^{-1}(4) \]

### Final Answer
The angle of projection \( \theta \) is:
\[ \theta = \tan^{-1}(4) \]