Aball thrown horizontall at 22.2 m/s from the roof of a building lands 36.0m from the base of the building. how tall is the building?

Velocity of the ball (v) = 22.2 m/s
Distance from the base of the building where the ball falls (d) = 36.0m
We know that, Velocity (v) = distance (d) / time (t)
From the above equation,
Time (t) = distance (d) / velocity (v)
Substituting the values for distance and velocity in the above equation,
Time = 36.0/22.2 = 1.62 seconds

When the ball is falling down, it accelerates due to gravity
This acceleration (a) = 9.8 m/s²

Now we need to find the height of the building.
The equation of motion for an object with constant acceleration is,
s = ut + 1/2 at²
Where,
s = vertical displacement, which is height of the building in our case
u = initial velocity of the ball = 0

Substituting all the above values in the equation of motion,
s = (0 × 1.62) + (1/2 × 9.8 × 1.62²)
s = 0 + (25.72/2) = 12.86 m = 12.9 m which is the height of the building

Therefore the building is 12.9m tall