An object is dropped from rest at a height of 128 m. Find the distance it falls during its final second in the air.

In the last second, the object traveled 45.6 m.

Explanation:

Hi there!

The equation of the height of the object at a time t 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.

t = time.

g = acceleration due to gravity (-9.8 m/s²).

First, let's find how much time it takes the object to reach the ground. For that, we have to find the value of "t" for which h = 0.

h = h0 + v0 · t + 1/2 · g · t²

Since the object is dropped and not thrown (v0 = 0).

h = h0 + 1/2 · g · t²

0 m  = 128 m - 1/2 · 9.8 m/s² · t²

-128 m / -4.9 m/s² = t²

t = 5.1 s

Now, let's find the height of the object 1 s before reaching the ground (at t = 4.1 s):

h = h0 + v0 · t + 1/2 · g · t²

h = 128 m - 1/2 · 9.8 m/s² · (4.1 s)²

h = 45.6 m

Then, in the last second (from 4.1 s to 5.1 s) the object traveled 45.6 m

hi there!

your question asks at which segment does freezing occur.

since we're talking about water, the freezing point of water will be 32 degrees fahrenheit or below.

hope this ! ~