Question

Question #f65ea

Answer 1

In terms of `h`:

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

In terms of `h` and `t`:

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

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")))" ")|)`