Problem: A particle is projected with a velocity u so that its horizontal range is twice the greatest attained. Find the horizontal range.

Solution:
To solve this problem, we need to use the equations of motion for projectile motion. Let's assume that the initial velocity of the particle is u and the angle of projection is θ.

Step 1: Find the time of flight
The time of flight of the particle can be found using the following equation:
T = 2usinθ/g
where u is the initial velocity, θ is the angle of projection, and g is the acceleration due to gravity.

Step 2: Find the maximum height reached
The maximum height reached by the particle can be found using the following equation:
H = u^2sin^2θ/2g

Step 3: Find the horizontal range
The horizontal range of the particle can be found using the following equation:
R = u^2sin2θ/g

Now, we are given that the horizontal range is twice the greatest attained. Therefore, we can write:
R = 2H

Substituting the values of H and R in the above equations, we get:
u^2sin^2θ/2g = 2u^2sin^2θ/g

Simplifying this equation, we get:
sin^2θ = 1/3

Therefore, sinθ = 1/√3

Substituting this value in the equation for horizontal range, we get:
R = u^2/2g

This is the final equation for the horizontal range of the particle.

Conclusion:
In conclusion, the horizontal range of a particle projected with a velocity u so that its horizontal range is twice the greatest attained is given by the equation R = u^2/2g, where u is the initial velocity and g is the acceleration due to gravity. The angle of projection can be found using the equation sinθ = 1/√3.

U^2/g as hrange=2hmaxu^2sin2theta/g= 2u^2/2gsin2theta=1sin2theta = π/2theta=π/4that comes out to be hrange =u^2/g