- #1

PhySci83

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- 0

## Homework Statement

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 1.6 s after it was tossed. What is the height of the power line and the initial speed of the ball?

## Homework Equations

y = y0 + v0 + 0.5*a*t^2

v^2 - V0^2 = 2*g*(y_final - y_initial)

## The Attempt at a Solution

To find when the ball was at the peak of the throw, I found the mid-point of 0.80 s and 1.6 s which is 1.2 s.

I calculated the maximum height of the ball by using the equation y = y_max + v0 + 0.5*a*t^2, where:

y = 0

y_max = desired

v0 = 0

a = -9.8 m/s^2

t = 1.2 s

Plugging these in, I got a maximum height of 7.056 m

Next, since I am looking for the height of the wire, I calculated the height difference between the maximum height of the path and the wire using the same equation, but using 0.4 s for time since from the height of the path to the ball passing the wire, it took 0.4 s. Again, I used the equation y = y_diff + v0 + 0.5*a*t^2, where:

y = 0

y_diff = desired

v0 = 0

a = -9.8 m/s^2

t = 0.4 s

Plugging these in, I got a differential height of 0.784 m.

To find the height of the line, I subtracted the difference from the maximum height:

7.056 m - 0.784 m = 6.272 m = 6.3 m (sig figs)

Then to find initial speed, I used the equation v^2 - v0^2 = 2*a*(y_final - y_initial), where:

v = 0

v_0 = desired

a = -9.8 m/s^2

y_final = 6.3 m

y_initial = 0

Plugging these in, I get an answer of 11.1 m/s.

I am submitting this into an online homework database and it tells me that my signs are wrong. I was under the assumption that speed and height are always positive since they are not vectors and it is only their counterparts, velocity and displacement, that carry the possible negative values.

Am I missing something?