If an object is dropped from a tall building and hits the ground 3.0 s later, how tall is the building?

s = 4.9t^2 = 4.9(9) = 44.1 m

To find the height of the building, you can use the formula for the displacement of a falling object:

h = (1/2) * g * t^2

Where:
h = height of the building
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time it takes for the object to hit the ground

Given that the object hits the ground 3.0 seconds later, we can substitute t = 3.0 seconds into the equation:

h = (1/2) * 9.8 * (3.0)^2

Calculating the equation gives us:

h = (1/2) * 9.8 * 9

h = 44.1 meters

Therefore, the building is approximately 44.1 meters tall.

To determine the height of the building, we can use the equation of motion for a falling object. The equation is given by:

h = 0.5 * g * t^2,

where h is the height of the building, g is the acceleration due to gravity (approximately 9.8 m/s^2 near the surface of the Earth), and t is the time it takes for the object to fall (3.0 s in this case).

Substituting the values into the equation, we get:

h = 0.5 * 9.8 * (3.0)^2

h = 0.5 * 9.8 * 9.0

h = 44.1 meters.

Therefore, the height of the building is approximately 44.1 meters.

30 feet