A player kicks up a ball at an angle `theta` to the horizontal. The horizontal range is maximum when `theta` equals-

Options

A. `30^(@)`
B. `45^(@)`
C. `60^(@)`
D. `90^(@)`

Answer

To find the angle \( \theta \) at which the horizontal range of a projectile is maximized, we can follow these steps:

### Step 1: Understand the Range Formula
The horizontal range \( R \) of a projectile launched at an angle \( \theta \) with an initial velocity \( v \) is given by the formula:
\[ R = \frac{v^2 \sin 2\theta}{g} \]
where \( g \) is the acceleration due to gravity.

### Step 2: Identify the Maximum Condition
To maximize the range \( R \), we need to maximize the term \( \sin 2\theta \). The sine function reaches its maximum value of 1.

### Step 3: Set Up the Equation
For \( \sin 2\theta \) to be equal to 1:
\[ \sin 2\theta = 1 \]

### Step 4: Solve for \( 2\theta \)
The sine function equals 1 at:
\[ 2\theta = 90^\circ + n \cdot 360^\circ \quad (n \in \mathbb{Z}) \]
For the first instance (n=0):
\[ 2\theta = 90^\circ \]

### Step 5: Solve for \( \theta \)
Dividing both sides by 2 gives:
\[ \theta = 45^\circ \]

### Conclusion
The angle \( \theta \) at which the horizontal range is maximum is:
\[ \theta = 45^\circ \]