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.
A tower is at a distance of `5m` from a man who can throw a stone with a maximum speed of `10m//s`. What is the maximum height that the man can hit on this tower.

Options

A. `4.5m`
B. `4m`
C. `5m`
D. `3.75m`

Explanation

To solve the problem, we need to find the maximum height that a stone can hit on a tower located 5 meters away from a man who can throw the stone with a maximum speed of 10 m/s. We will use the trajectory equation provided:

\[ y = x \tan \theta - \frac{g x^2}{2 u^2 \cos^2 \theta} \]

Where:
- \( y \) is the height,
- \( x \) is the horizontal distance (5 m),
- \( \theta \) is the angle of projection,
...