Question

[Motion in a line] The superhero Green Lantern steps from the top of a tall building.
He falls freely from rest to the ground, falling half the total distance
to the ground during the last 1.00 s of his fall. What is the
height h of the building?

Answer: H=57.1 m
I'm still not sure how to solve this though, maybe someone can help?

Answer 1

Start with the formula for distance travelled under uniform acceleration.
\[s=ut+\frac{ 1 }{ 2 }at^2\]
You can apply this to the entire fall, and you can also apply it to the fall from the top to the point halfway down. You know that the second case takes one second less than the first case. (u is zero in both cases)
This gives you two equations for the total height and total time.
I recommend you solve them for the total time, then substitute back to find the total height.

Answer 2

I don't think you DO know that the second half is 1 second less.

What you DO know is that s is the same in both cases. u is not 0 for the second half - he is already falling fast

Answer 3

I am saying that the first half is one second less than the whole fall.

Answer 4

Because the second half of the fall takes precisely one second, as stated in the question.

Answer 5

you know that the time for the second half is 1s

so s= u +0.5g (for the second half)

Answer 6

I could put my equations up in detail but i thought you might like to play around with it yourself first

Answer 7

for the first half

v^2 = 2 as

and v for the first half is u for the second half

so s= sgrt(2as)+0.5a

Answer 8

wait guys, what does u stand for first? uniform acceleration? if a is acceleration, which is 9.8 m/s² in this case, what's u exactly?

Answer 9

u is the initial velocity

Answer 10

u is the initial velocity
at the top this is 0

but at the start of the second half he has a speed equal to "v" (the final vleocit)for the first half

Answer 11

mrnood your approach looks fine to me, you do still have to solve your equation for s to get the final answer

Answer 12

the initial velocity in the first part's 0 because the hero was at rest, in the second time the time difference between t0 ( start of second part) and tf (final time) is 1 second. so i can use that in equation right? like
a(t) = g
v(t) = g(tf-t0) + v(t0)
x(t) = g(tf-t0)^2 + v(t0)(tf-t0) + x(t0)

Answer 13

my initial velocity is always 0 because i consider in one case the entire motion, and in the second case only the first half of the motion - u=0 in both my cases

Answer 14

yeah - solving that is not a simple as I expected....

Answer 15

so you separate them? and then add both s's of the first and second equation to end up with h? well h is 2x0 after all, i'm confused with it though. i don't know why.

Answer 16

suppose the total time to fall is T and the total height is H
then for the whole drop we have \[H = \frac{ 1 }{ 2 }gT^2\]
call this equation 1

Answer 17

Now consider only the first half of the drop (half in distance i mean) then we have\[\frac{ H }{ 2}=\frac{ 1 }{ 2}g(T-1)^2\] call this equation 2
because it takes T-1 seconds to fall the first H/2 from rest

Answer 18

Now you can eliminate H from equations 1 and 2, you get a quadratic in T, only one of the two possible values of T will make sense

Answer 19

Then you can substitute that value of T in equation 1 to calculate H

Answer 20

really ? it can be solved like that? well i aske a friend, he said an easier way is to do;
delta integral vx dt = alpha integral vx dt
0 -> t0 t0 -> t0+1
where alpha is the ratio between the two. well that's a solution, another solution i looked up is :

Answer 21

First half of distance:
2g(h/2) = v²
v = sqrt(gh)

Second half of distance:
h/2 = vt + gt²/2
h/2 = t sqrt(gh) + gt²/2
h - 2t sqrt(gh) - gt² = 0

sqrt(h) = [2t sqrt(g) + sqrt{4gt² + 4gt²}] / 2 (can't be negative)
sqrt(h) = t[1 + sqrt(2)] sqrt(g)
h = gt²[1 + sqrt(2)]²
h = (9.8 m/s²)(1.00 s)²[1 + sqrt(2)]²
h = 57.1 m

he kinda lost me at sqrt(h) = [2t sqrt(g) + sqrt{4gt² + 4gt²}] / 2 (can't be negative)
i dunno where that came from, exactly.

Answer 22

execuse me ?? means ^2.

Answer 23

2g(h/2) = v^2
v = sqrt(gh)

Second half of distance:
h/2 = vt + gt^2/2
h/2 = t sqrt(gh) + gt^2/2
h - 2t sqrt(gh) - gt^2 = 0

sqrt(h) = [2t sqrt(g) + sqrt{4gt^2 + 4gt^2}] / 2 (can't be negative)
sqrt(h) = t[1 + sqrt(2)] sqrt(g)
h = gt^2[1 + sqrt(2)]^2
h = (9.8 m/s^2)(1.00 s)^2[1 + sqrt(2)]^2
h = 57.1 m

Answer 24

yes this approach is fine, it's what mrnood was suggesting also - but can you produce the answer for yourself ? that is the important point - good luck

Answer 25

yeah, thanks. to be honest i just wanted some clearing out, i never considered t to be a duration between two points, (t and t0) it was always instant for me, so this came with confusion.

Answer 26

the question is intended to give you practice with the equations for motion in a straight line under constant acceleration, bear that in mind, keep a clear head and play around with it yourself : )

Answer 27

yeah, well.. i've been doing that. i've made some computational errors and other conceptional errors and got irritated, which is why i asked for help. so thanks i'll go at it again, but now i know how to solve it too. xD