The vertical height of the projectile at time t is given by y = 4t – 5t^{2} and the horizontal distance covered is given by x = 3t. What is the angle of projection with the horizontal?

(A) tan^{–1} 3/5

(B) tan^{–1} 4/5

(C) tan^{–1} 4/3

(D) tan^{–1} 3/4