# A bird is sitting on a power line. It defecates. How fast is the secretion going after 2 seconds?

Jun 21, 2017

$\text{speed" = 19.6"m"/"s}$

#### Explanation:

I'll assume it is still in the air after $2$ $\text{s}$..

To solve this problem, we can use the formula

${v}_{y} = {v}_{0 y} + {a}_{y} t$

where

• ${v}_{y}$ is the velocity of the secretion at time $t = 2$ $\text{s}$

• ${v}_{0 y}$ is the initial velocity of the secretion, which I'll assume is $0$ for these purposes, although it may have required some pushing to "get it out," and would then actually have some initial speed

• ${a}_{y}$ is the acceleration, which is simply $- g$, which is $- 9.8$ "m"/("s"^2)

• $t$ is the time, which is given as $2$ $\text{s}$

Plugging in the known variables, we have

v_y = 0"m"/"s" + (-9.8"m"/("s"^2))(2color(white)(l)"s")

v_y = color(red)(-19.6"m"/"s"

This is its velocity, which includes a negative sign to indicate direction (it's traveling downward). If you asked just how fast it is going, then its speed is just the positive quantity

$\textcolor{red}{19.6 \text{m"/"s}}$

(or, if you decide to use only one significant figure, $20 \text{m"/"s}$).

The speed of the bird's feces after $2$ seconds is thus color(red)(19.6 sfcolor(red)("meters per second".

Extra info :

Just for information, after $2$ $\text{s}$ and with assumedly no initial speed, the secretion would have traveled

$\Delta y = {\left(0 \text{m"/"s")(2color(white)(l)"s") + 1/2(-9.8"m"/("s"^2))(2color(white)(l)"s}\right)}^{2} = - 19.6$ $\text{m}$

$19.6$ meters downward, so the height of the wire the bird was sitting on must have been AT LEAST $19.6$ $\text{m}$ high, which is a little higher than most standard telephone poles (which are$\approx 10$ $\text{m}$ tall)..

Some taller poles can reach heights of about $37$ meters, so he could have been sitting on one of these...interesting question by the way!