When a projectile is launched at an angle with the horizontal, it follows a curved path called a trajectory. The angle of projection is the angle between the horizontal and the initial velocity of the projectile. In this article, we will discuss how to find the angle of projection for maximum horizontal range and the corresponding height achieved by the projectile.


The maximum horizontal range is the farthest distance that a projectile can travel horizontally. It is achieved when the projectile is launched at a certain angle of projection. To find this angle, we need to use the following formula:

θ = tan⁻¹(gR/v²)

Where θ is the angle of projection, g is the acceleration due to gravity, R is the horizontal range, and v is the initial velocity of the projectile.


To find the corresponding height achieved by the projectile, we need to use the following formula:

H = (v²sin²θ)/(2g)

Where H is the maximum height achieved by the projectile.


Suppose a projectile is launched with an initial velocity of 20 m/s at an angle of 30° with the horizontal. Find the angle of projection for its maximum horizontal range and the corresponding height achieved by the projectile.

Using the formula, we have:

θ = tan⁻¹(gR/v²)
θ = tan⁻¹(9.8 x R/400)

To find the value of R, we can use the formula:

R = (v²sin2θ)/g
R = (20²sin60°)/9.8
R = 40.82 m

Substituting the value of R, we have:

θ = tan⁻¹(9.8 x 40.82/400)
θ = 24.06°

Therefore, the angle of projection for maximum horizontal range is 24.06°.

To find the corresponding height achieved, we have:

H = (v²sin²θ)/(2g)
H = (20²sin²30°)/(2 x 9.8)
H = 20.41 m

Therefore, the corresponding height achieved by the projectile is 20.41 m.


In conclusion, the angle of projection for maximum horizontal range and the corresponding height achieved by the projectile can be found using the formulas mentioned above. It is important to understand these concepts in order to accurately predict the trajectory of a projectile.

Theta must be 45 degree!