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 step by step, we will use the equations of motion for projectile motion and the relationships between the components of the initial velocity.

### Step 1: Understand the relationships
We know that:
- The maximum height \( h \) of a projectile is given by:
\[ h = \frac{u_y^2}{2g} \]
where \( u_y = u \sin \theta \) is the vertical component of the initial velocity and \( g \) is the acceleration due to gravity.

- The horizontal range \( R \) of a projectile is given by:
\[ R = \frac{u_x \cdot u_y}{g} \]
where \( u_x = u \cos \theta \) is the horizontal component of the initial velocity.

### Step 2: Set up the relationship from the problem statement
According to the problem, the maximum height attained by the projectile is half of the horizontal range:
\[ h = \frac{1}{2} R \]

### Step 3: Substitute the expressions for \( h \) and \( R \)
Substituting the expressions for \( h \) and \( R \) into the equation:
\[ \frac{u_y^2}{2g} = \frac{1}{2} \left( \frac{u_x \cdot u_y}{g} \right) \]

### Step 4: Simplify the equation
Multiplying both sides by \( 2g \) to eliminate the denominators:
\[ u_y^2 = \frac{1}{2} u_x \cdot u_y \]

### Step 5: Rearranging the equation
We can rearrange this equation:
\[ 2u_y^2 = u_x \cdot u_y \]
Assuming \( u_y \neq 0 \), we can divide both sides by \( u_y \):
\[ 2u_y = u_x \]

### Step 6: Substitute the components of velocity
Substituting \( u_y = u \sin \theta \) and \( u_x = u \cos \theta \):
\[ 2(u \sin \theta) = u \cos \theta \]

### Step 7: Cancel \( u \) (assuming \( u \neq 0 \))
Dividing both sides by \( u \):
\[ 2 \sin \theta = \cos \theta \]

### Step 8: Rearranging to find \( \tan \theta \)
Dividing both sides by \( \cos \theta \):
\[ 2 \tan \theta = 1 \]
Thus,
\[ \tan \theta = \frac{1}{2} \]

### Step 9: Find \( \theta \)
Taking the inverse tangent:
\[ \theta = \tan^{-1}\left(\frac{1}{2}\right) \]

### Conclusion
Thus, the angle \( \theta \) is equal to \( \tan^{-1}(2) \).