Given data:
* The initial velcoity of the ball is 40 m/s.
* The angle made by the velocity with the horizontal is 30 degree.
Solution:
The horizontal range of the projectile in terms of the initial velocity is,

where R is the horizontal range, u is the initial velocity, g is the acceleration due to gravity, and

Substituting the known values,

Thus, the ball travel 141.4 m in the horizontal range.