The maximum height attained by a projectile when thrown at an angle `theta` with the horizontal is found to be half the horizontal range. Then `theta` is equal to

Options

A. `tan^(-1) (2)`
B. `pi/6`
C. `pi/4`
D. `tan^(-1)(1/2)`

Answer

To solve the problem, we need to find the angle \( \theta \) at which a projectile is thrown such that the maximum height attained is half of the horizontal range. Let's break this down step by step.

### Step 1: Understand the equations for projectile motion
The maximum height \( H \) of a projectile thrown at an angle \( \theta \) with an initial velocity \( u \) is given by the formula:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]
where \( g \) is the acceleration due to gravity.

The horizontal range \( R \) of the projectile is given by:
\[ R = \frac{u^2 \sin 2\theta}{g} \]

### Step 2: Set up the relationship given in the problem
According to the problem, the maximum height \( H \) is half of the horizontal range \( R \):
\[ H = \frac{1}{2} R \]

### Step 3: Substitute the formulas for \( H \) and \( R \)
Substituting the expressions for \( H \) and \( R \) into the equation:
\[ \frac{u^2 \sin^2 \theta}{2g} = \frac{1}{2} \left( \frac{u^2 \sin 2\theta}{g} \right) \]

### Step 4: Simplify the equation
Cancelling \( \frac{u^2}{g} \) from both sides (assuming \( u \neq 0 \)):
\[ \frac{\sin^2 \theta}{2} = \frac{1}{2} \sin 2\theta \]

### Step 5: Use the identity for \( \sin 2\theta \)
Recall that \( \sin 2\theta = 2 \sin \theta \cos \theta \). Thus, we can rewrite the equation:
\[ \frac{\sin^2 \theta}{2} = \frac{1}{2} (2 \sin \theta \cos \theta) \]
This simplifies to:
\[ \sin^2 \theta = 2 \sin \theta \cos \theta \]

### Step 6: Rearrange the equation
Rearranging gives:
\[ \sin^2 \theta - 2 \sin \theta \cos \theta = 0 \]
Factoring out \( \sin \theta \):
\[ \sin \theta (\sin \theta - 2 \cos \theta) = 0 \]

### Step 7: Solve for \( \theta \)
This gives us two cases:
1. \( \sin \theta = 0 \) (which corresponds to \( \theta = 0^\circ \) or \( 180^\circ \), not valid for projectile motion)
2. \( \sin \theta - 2 \cos \theta = 0 \) or \( \sin \theta = 2 \cos \theta \)

Dividing both sides by \( \cos \theta \):
\[ \tan \theta = 2 \]

### Step 8: Find \( \theta \)
Taking the inverse tangent:
\[ \theta = \tan^{-1}(2) \]

### Final Answer
Thus, the angle \( \theta \) is:
\[ \theta = \tan^{-1}(2) \]