1. Dropping object from height

Hello,
An object is released from a height of 25M. How far has the object travelled during the 3rd second of fall?

I don't really know how to do this - what formula/'s should I use?

thanks!

2. Formula is
$h = v_0 \cdot t + a\frac{t^2}{2}$
$0 = v_0$
$a = g$
$h = g\frac{t^2}{2}$
$-25 = g\frac{t^2}{2}$
You can find that object touches the ground at 2.26 seconds
Find how far it went at 2 seconds and do 25m minus this answer

The formula is

$h = ut +\frac{1}{2}gt^2$

where h = height

u = initial velocity (which is 0 here)

g = accleration due to gravity $= 9.81 \;m/s^2$

t = time.

4. Got it - thanks a lot! Oddly worded question, that was the tough part.

5. Don't you think the object should travel further during the 3rd second.
It starts from heigth 25 m. And goes down 19 m in two seconds as you said.
It is going to make the rest (6m) in the third second.

6. Originally Posted by vincisonfire
Don't you think the object should travel further during the 3rd second.
It starts from heigth 25 m. And goes down 19 m in two seconds as you said.
It is going to make the rest (6m) in the third second.
Height fallen by object after 2 seconds is given by

$h = 0 + \frac{1}{2}(9.81)(2)^2$

$h=19.62\;m$

And, Height fallen by object after 3 seconds is given by

$h = 0 + \frac{1}{2}(9.81)(3)^2$

$h=44.15\;m>25\;m$

so, object touches ground before 3 seconds

Height fallen by object in 3rd second = 25 - 19.62 = 5.38 m

7. it only takes a bit over 2.2 seconds to hit the ground from a height of 25 m, so asking how far it travels in the third second is poorly posed.

if it had enough height to fall the "third" second, then from t = 2 to t = 3 seconds ...

$\Delta y = v(2) \cdot \Delta t - \frac{1}{2}g(\Delta t)^2 = -19.6(1) - 4.9(1) = -24.5 \, \, m$

if the intent was to determine the distance it fell from t = 2 seconds until it hits the ground, ...

$\Delta y = -\frac{1}{2}g(2)^2 = -19.6 \, \, m$

leaving 5.4 m to fall from t = 2 to impact.