What formulas do I use for this?

Since the problem does not mention the angle of the brick’s initial velocity, let’s assume the brick’s initial velocity is horizontal. Use the following equation to determine the height of the building.

d = vi * t + ½ * a * t^2

vi = initial vertical velocity = 0 m/s, a = 9.8 m/s^2

d = ½ * 9.8 * 4^2 = 78.4 meters

Use the following equation to determine the horizontal distance from the base of the building to the place where the brick landed.

d = v * t = 2 * 4 = 8 meters

Use whatever equation you can derive by knowing the physics. What I'm saying is you're not learning physics by memorizing equations.

So here's the physics you should learn.

The definition for distance = average speed X time traveled; S = Vavg T is the first derivation. Note equations are just shorthand for what we can say in narrative..."distance equals average speed time time traveled."

Now some physics, the velocity V after T time of acceleration A, is the initial velocity U plus that acceleration over time; V = U + AT is the second derivation.

And more physics velocity V is a vector and consists of X and Y components Vx and Vy. So they form the right triangle Vx, Vy, V. Which means V = sqrt(Vx^2 + Vy^2) is the magnitude (speed) from the Pythagorean.

Now you have the physics, solve the problem by starting with the givens. You have Ux = Vx = 2 mps. And T = 4 s. You are looking for Sy = Vyavg T and Sx = Vxavg T the two distances, Y and X direction.

As we assume no air resistance, Ux = Vx = Vxavg = 2 mps is constant speed. So Sx = Vxavg T = 2*4 = 8 m from the base ANS.

More physics...a body in free fall, like that brick, will accelerate at Ay = g = 9.8 m/s^2 near Earth's service. And because that's a constant acceleration we can show that Vyavg = (Vy + Uy)/2; where Uy = 0 is the initial vertical speed from the given, pushed off.

So we have Sy = Vyavg T = Vy/2 * T and as Vy = AyT = gT that derives into Sy = 1/2 gT^2 = 4.9*16 = 78.4 m building height. ANS.

The fall motion is parabolic horizontal (no slope angle).

initial speed u = 2m /s

time of hits = 4s

The drop height h = 1/2 gt ^ 2

horizontal distance from the base of the building Dx = u * t (t = fall time)