Answer:
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
