How long would it take an object to reach the ground from the top of a building that is 470 feet tall? round to the nearest tenth of a second.

It would take the object 5.4 s to reach the ground.

Explanation:

Hi there!

The equation of the height of a free-falling object at any given time, neglecting air resistance, is the following:

h = h0 + v0 · t + 1/2 · g · t²

Where:

h = height of the object at time t.

h0 = initial height.

v0 = initial velocity.

g = acceleration due to gravity (-32.2 ft/s² considering the upward direction as positive).

t = time

Let´s supose that the object is dropped and not thrown so that v0 = 0. Then:

h = h0 + 1/2 · g · t²

We have to find the time at which h = 0:

0 = 470 ft - 1/2 · 32.2 ft/s² · t²

Solving for t:

-470 ft = -16.1 ft/s² · t²

-470 ft / -16.1 ft/s² = t²

t = 5.4 s

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.