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.