The trajectory of a projectile is given by `y=x tantheta-(1)/(2)(gx^(2))/(u^(2)cos^(2)theta)`. This equation can be used for calculating various phenomen such as finding the minimum velocity required to make a stone reach a certain point maximum range for a given projection velocity and the angle of projection required for maximum range. The range of a particle thrown from a tower is define as the distance the root of the tower and the point of landing.
In the previous problem, what should be the corresponding projection angle.

Options

A. `tan^(-1)(1//2)`
B. `tan^(-1)(1//3)`
C. `tan^(-1)2`
D. `tan^(-1)3`

Explanation

To solve the problem of finding the corresponding projection angle for the trajectory of a projectile thrown from a tower, we can follow these steps:

### Step 1: Understand the Given Trajectory Equation
The trajectory of a projectile is given by the equation:
\[ y = x \tan \theta - \frac{1}{2} \frac{g x^2}{u^2 \cos^2 \theta} \]
where:
- \( y \) is the height,
- \( x \) is the horizontal distance,
...