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.
From a certain tower of unknown height it is found that the maximum range at a certain projection velocity is obtained for a projection angle of `30^(@)` and this range is `10sqrt(3)m`. The projection velocity must be

Options

A. `10m//s`
B. `10sqrt(3)m//s`
C. `(10)/(sqrt(3))m//s`
D. `5m//s`

Explanation

To solve the problem, we need to find the projection velocity \( u \) of a projectile thrown from a tower, given that the maximum range at a projection angle of \( 30^\circ \) is \( 10\sqrt{3} \, m \).

### Step-by-Step Solution:

1. **Understanding the Range Formula**:
The range \( R \) of a projectile launched from a height can be expressed as:
\[
R = \frac{u^2 \sin(2\theta)}{g} + \text{(height effect)}
...