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 the maximum height attained by a projectile is half of its horizontal range.

### Step 1: Write down the formulas for maximum height and range.
The maximum height \( H \) of a projectile thrown at an angle \( \theta \) with initial velocity \( u \) is given by:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]
The horizontal range \( R \) of the projectile is given by:
\[ R = \frac{u^2 \sin 2\theta}{g} \]

### Step 2: Set up the relationship between maximum height and range.
According to the problem, the maximum height is half of the horizontal range:
\[ H = \frac{1}{2} R \]
Substituting the formulas for \( H \) and \( R \):
\[ \frac{u^2 \sin^2 \theta}{2g} = \frac{1}{2} \left( \frac{u^2 \sin 2\theta}{g} \right) \]

### Step 3: Simplify the equation.
Multiplying both sides by \( 2g \) to eliminate the denominators:
\[ u^2 \sin^2 \theta = u^2 \sin 2\theta \]
Now, we can cancel \( u^2 \) from both sides (assuming \( u \neq 0 \)):
\[ \sin^2 \theta = \sin 2\theta \]

### Step 4: Use the double angle identity.
Recall that \( \sin 2\theta = 2 \sin \theta \cos \theta \). Therefore, we can rewrite the equation:
\[ \sin^2 \theta = 2 \sin \theta \cos \theta \]

### Step 5: 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 6: Solve for \( \theta \).
This gives us two cases:
1. \( \sin \theta = 0 \) which implies \( \theta = 0^\circ \) (not a valid angle for projectile motion).
2. \( \sin \theta - 2 \cos \theta = 0 \) which implies:
\[ \sin \theta = 2 \cos \theta \]
Dividing both sides by \( \cos \theta \) (assuming \( \cos \theta \neq 0 \)):
\[ \tan \theta = 2 \]
Therefore, \( \theta = \tan^{-1}(2) \).

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