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

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!