Question #f65ea
1 Answer
In terms of
color(red)(t_"final meter" = (sqrt2color(white)(l)"m")/(sqrt(g(h-1color(white)(l)"m")) + sqrt(gh))tfinal meter=√2lm√g(h−1lm)+√gh
In terms of
color(blue)(t_"final meter" = ((2color(white)(l)"m")t)/(2h + tsqrt(2g(h-1color(white)(l)"m")))tfinal meter=(2lm)t2h+t√2g(h−1lm)
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
(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
Now what we can do is plug these two expressions in for
Deltay = ((sqrt(2g(h-1color(white)(l)"m")) + sqrt(2gh))/2)t
Now, we plug in the distance
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
We have two situations we're trying to relate: the motion as it falls from the height
We can recognize that the final velocity (which I'll call
With that in mind, let's find an equation that involves the final velocity when it's dropped from the height
h = ((v_0 + v_2)/2)t
But since it's dropped from rest, the initial velocity
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 = (2color(white)(l)"m")/(t_"final meter") - v_1
Here, the final velocity
Recall from earlier that we found this velocity
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
(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 (
(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")))" ")|)