Question #2e587

We're asked to find the time, in seconds, it takes an object to fall to Earth's surface from a height of #250# #"m"#.

To do this, we can use the kinematics equation

#ul(y = y_0 + v_(0y)t - 1/2g t^2#

where

• #y# is the height at time #t# (which is #0#, ground-level)

• #y_0# is the initial height (given as #250# #"m"#)

• #v_(0y)# is the initial velocity (it dropped from a state of rest, so this is #0#)

• #t# is the time (what we're trying to find)

• #g = 9.81# #"m/s"^2#

Sine the initial #y#-velocity is #0#, we can change the equation to

#y = y_0 - 1/2g t^2#

Let's solve this for our unknown variable, #t#:

#y-y_0= -1/2g t^2#

#-2(y-y_0) = g t^2#

#t^2 = (-2(y-y_0))/g#

#color(red)(t = sqrt((-2(y-y_0))/g)#

Plugging in known values:

#t = sqrt((-2(0-250color(white)(l)"m"))/(9.81color(white)(l)"m/s"^2)) = color(blue)(ulbar(|stackrel(" ")(" "7.14color(white)(l)"s"" ")|)#

So ultimately, if you're ever given a situation where you're asked to find the time it takes an object to fall a certain distance (with #0# initial velocity), just use the simplified equation

#color(red)(t = sqrt((2*"height")/g)#