**Answer:**

a) The height of the power line is 4.7 m

b) The initial velocity is 9.8 m/s

**Explanation:**

The equation for the height of the ball is the following:

y = y0 + v0 · t + 1/2 · g · t²

Where:

y = height of the ball

y0 = initial height

v0 = initial velocity

t = time

g = acceleration due to gravity

Let´s place the origin of the frame of reference at the throwing point so that y0 = 0

b) We know that when the time is 0.80 s and 1.2 s the height is the same. Then:

y = y0 + v0 · t + 1/2 · g · t²

y = v0 · 0.80 s - 1/2 · 9.8 m/s² · (0.80 s)²

y = v0 · 1.2 s - 1/2 · 9.8 m/s² · (1.2 s)²

v0 · 0.80 s - 1/2 · 9.8 m/s² · (0.80 s)² = v0 · 1.2 s - 1/2 · 9.8 m/s² · (1.2 s)²

v0 · 0.80 s - 3.136 m = v0 · 1.2 s - 7.056 m

v0 · 0.80 s - v0 · 1.2 s = - 7.056 m + 3.136 m

-0.4 s · v0 = -3.92 m

v0 = -3.92 m / -0.4 s

v0 = 9.8 m/s

The initial velocity is 9.8 m/s

a) Now, we can calculate the height of the power line:

y = y0 + v0 · t + 1/2 · g · t²

y = 9.8 m/s · 1.2 s - 1/2 · 9.8 m/s² · (1.2 s)²

y = 4.7 m

The height of the power line is 4.7 m