Question

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

Answer 1

`"speed" = 19.6"m"/"s"`

Explanation:

I'll assume it is still in the air after `2` `"s"`..

To solve this problem, we can use the formula

`v_y = v_(0y) + a_yt`

where

• `v_y` is the velocity of the secretion at time `t = 2` `"s"`

• `v_(0y)` 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` `"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

`color(red)(19.6"m"/"s")`

(or, if you decide to use only one significant figure, `20"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` `"s"` and with assumedly no initial speed, the secretion would have traveled

`Deltay = (0"m"/"s")(2color(white)(l)"s") + 1/2(-9.8"m"/("s"^2))(2color(white)(l)"s")^2 = -19.6` `"m"`

`19.6` meters downward, so the height of the wire the bird was sitting on must have been AT LEAST `19.6` `"m"` high, which is a little higher than most standard telephone poles (which are`~~10` `"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!