Question #f65ea

1 Answer
Sep 2, 2017

In terms of hh:

color(red)(t_"final meter" = (sqrt2color(white)(l)"m")/(sqrt(g(h-1color(white)(l)"m")) + sqrt(gh))tfinal meter=2lmg(h1lm)+gh

In terms of hh and tt:

color(blue)(t_"final meter" = ((2color(white)(l)"m")t)/(2h + tsqrt(2g(h-1color(white)(l)"m")))tfinal meter=(2lm)t2h+t2g(h1lm)

Explanation:

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")))" ")|)