To find the height of an overhead power line, you throw a ball straight upward. the ball passes the line on the way up after 0.80 s, and passes it again on the way down 1.2 s after it was tossed. (a) what is the height of the power line? (b) what is the initial speed of the ball?

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

what do the notation's look like?

3. dark matter

empty space could also be a simple vacuum