A brick is released with no initial speed from the roof of a building and strikes the ground in 1.50 {s}, encountering no appreciable air drag.

A. How tall, in meters, is the building?

B.How fast is the brick moving just before it reaches the ground?

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To find the height of the building and the velocity of the brick just before it hits the ground, we can use the kinematic equations of motion. In this case, we will use the equation:

\(h = \frac{1}{2}gt^{2}\) (Equation 1)

where h is the height of the building (in meters), g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time taken for the brick to hit the ground (in seconds).

For part A, we can rearrange Equation 1 to solve for h:
\(h = \frac{1}{2}gt^{2}\)

Now, we substitute the given values into the equation:
\(h = \frac{1}{2}(9.8)(1.50)^{2}\)

Simplifying the equation:
\(h = 1.47\) meters

Therefore, the height of the building is approximately 1.47 meters.

For part B, we can use the equation:

\(v = gt\) (Equation 2)

where v is the velocity of the brick (in m/s).

To find the velocity of the brick just before it reaches the ground, we substitute the value of t into Equation 2:
\(v = (9.8)(1.50)\)

Simplifying the equation:
\(v = 14.7\) m/s

Therefore, the velocity of the brick just before it hits the ground is approximately 14.7 m/s.