Question #f65ea

We're asked to find the time it takes a body in free-fall to travel the final meter of its motion.

I'll assume the body starts from rest.

To do this, we can first use the kinematics equation

#Deltay = ((v_(0y) + v_y)/2)t#

where

• #Deltay# is the distance it falls (#1# #"m"#)

• #v_(0y)# is the initial velocity of the body

• #v_y# is the final velocity of the body

• #t# is the time it takes to travel the final meter

We need to find the initial and final velocities of the body, for which we can use the equation

#ul((v_y)^2 = (v_(0y))^2 + 2gh#

Or, since the initial #y#-velocity is #0#,

#(v_y)^2 = 2gh#

or

#v_y = sqrt(2gh)#

We can make two equations that model this one for the final velocity and the initial velocity; we then have

#v_y = sqrt(2gh)" "# (final velocity)

#v_(0y) = sqrt(2g(h-1color(white)(l)"m"))" "# (initial velocity)

The initial velocity was obtained by realizing that it is the equivalent of the final velocity, but if the body were dropped from a height of #1# #"m"# lower (i.e. #h-1#, the height minus one meter).

Now what we can do is plug these two expressions in for #v_y# and #v_(0y)# in our first equation:

#Deltay = ((sqrt(2g(h-1color(white)(l)"m")) + sqrt(2gh))/2)t#

Now, we plug in the distance #Deltay# as #1# #"m"#, and solve for the time it takes to fall that distance, #t#:

#1color(white)(l)"m" = ((sqrt(2g(h-1color(white)(l)"m")) + sqrt(2gh))/2)t#

#2color(white)(l)"m" = (sqrt(2g(h-1color(white)(l)"m")) + sqrt(2gh))t#

#t = (2color(white)(l)"m")/(sqrt(2g(h-1color(white)(l)"m")) + sqrt(2gh))#

Or

#color(red)(ulbar(|stackrel(" ")(" "t = (sqrt2color(white)(l)"m")/(sqrt(g(h-1color(white)(l)"m")) + sqrt(gh))" ")|)#

This equation is used in terms of only the height #h#. If you want the value for the time in terms of this height and the time it takes to reach the ground from that height, here's the solution:

We have two situations we're trying to relate: the motion as it falls from the height #h#, and the motion it falls in the final #1# #"m"# of its motion.

We can recognize that the final velocity (which I'll call #v_2#) for both situations will be the same; i.e. it strikes the ground with the same speed.

With that in mind, let's find an equation that involves the final velocity when it's dropped from the height #h#. We can again use the equation

#h = ((v_0 + v_2)/2)t#

But since it's dropped from rest, the initial velocity #v_0 = 0#:

#h = ((v_2)/2)t#

Solving for the final velocity:

#color(green)(v_2 = (2h)/(t)#

Now let's find the final velocity equation relating the motion in the final meter, using the same equation yet again:

#1color(white)(l)"m" = ((v_1 + v_2)/2)xxoverbrace(t_"final meter")^"time taken during final meter"#

Solving for #v_2#:

#v_2 = (2color(white)(l)"m")/(t_"final meter") - v_1#

Here, the final velocity #v_2# is the same that we just found, and #v_1# is the initial velocity as it enters the final meter.

Recall from earlier that we found this velocity #v_1# to be

#v_1 = sqrt(2g(h-1color(white)(l)"m"))#

(this is the same as #v_(0y)# from above)

Substituting this into the last equation, we have

#color(green)(v_2 = (2color(white)(l)"m")/(t_"final meter") - sqrt(2g(h-1color(white)(l)"m"))#

Now, let's set the two #color(green)("green"# equations equal to each other:

#(2h)/t = (2color(white)(l)"m")/(t_"final meter") - sqrt(2g(h-1color(white)(l)"m"))#

Now all we need to do is solve for the time it takes to travel the final meter of motion (#t_"final meter"#):

#(2color(white)(l)"m")/(t_"final meter") = (2h)/t + sqrt(2g(h-1color(white)(l)"m"))#

#t_"final meter" = (2color(white)(l)"m")/((2h)/t + sqrt(2g(h-1color(white)(l)"m")))#

Or

#color(blue)(ulbar(|stackrel(" ")(" "t_"final meter" = ((2color(white)(l)"m")t)/(2h + tsqrt(2g(h-1color(white)(l)"m")))" ")|)#