A ball is projected with speed u at an angle `theta` to the horizontal. The range R of the projectile is given by
`R=(u^(2) sin 2theta)/(g)`
for which value of `theta` will the range be maximum for a given speed of projection? (Here g= constant)

Options

A. `pi/2 rad`
B. `pi/4 rad`
C. `pi/3 rad`
D. `pi/6 rad`

Explanation

To find the angle \( \theta \) at which the range \( R \) of a projectile is maximized, we start with the formula for the range:

\[ R = \frac{u^2 \sin 2\theta}{g} \]

where:
- \( R \) is the range,
- \( u \) is the initial speed of projection,
- \( \theta \) is the angle of projection,
- \( g \) is the acceleration due to gravity (a constant).

### Step 1: Understand the function to maximize
We need to maximize \( R \) with respect to \( \theta \). Since \( u \) and \( g \) are constants, we can focus on maximizing \( \sin 2\theta \).

### Step 2: Find the maximum of \( \sin 2\theta \)
The sine function reaches its maximum value of 1. Therefore, we set:

\[ \sin 2\theta = 1 \]

### Step 3: Solve for \( 2\theta \)
The sine function equals 1 at:

\[ 2\theta = \frac{\pi}{2} + 2n\pi \quad (n \in \mathbb{Z}) \]

For the first occurrence (the smallest positive angle), we take \( n = 0 \):

\[ 2\theta = \frac{\pi}{2} \]

### Step 4: Solve for \( \theta \)
Dividing both sides by 2 gives:

\[ \theta = \frac{\pi}{4} \]

### Conclusion
Thus, the angle \( \theta \) that maximizes the range \( R \) for a given speed of projection \( u \) is:

\[ \theta = \frac{\pi}{4} \text{ radians} \]