How far below an initial straight-line path will a projectile fall in three seconds?

If the particle was on some diagonal straight line path ,so in three seconds it will have gone a horizontal distance, Xa = Vx(3) and a vertical height, Ya = Vy(3) , then because gravity is pulling it down it will drop a distance Y = (1/2)(9.8)(3^2) = 44.1 m

So the distance below Ya the projectile falls is always 44.1 m (in 3 sec), regardless of the initial speed or angle of launch (both of which are related to Vx and Vy but do not effect the amt. of drop due to gravity)

To look at it from the kinematic eqs. The X and Y position of a projectile with velocity components Vx and Vy are;

X = (Vx)t

Y = (Vy)t  -  (1/2)(g)(t^2)

The X eq. tells you how far it goes horizontally in time "t".

The Y eq , which tells you how high it goes in the same time "t", has two parts
The first part, (Vy)t , tells you how high it would go if there were no gravity.
The second part tells you how far it drops because there is gravity.