# Dropping object from height

• Oct 22nd 2008, 05:05 PM
jeepwheeler
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!
• Oct 22nd 2008, 05:26 PM
vincisonfire
Formula is
$\displaystyle h = v_0 \cdot t + a\frac{t^2}{2}$
$\displaystyle 0 = v_0$
$\displaystyle a = g$
$\displaystyle h = g\frac{t^2}{2}$
$\displaystyle -25 = g\frac{t^2}{2}$
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
• Oct 22nd 2008, 05:32 PM
Shyam
The formula is

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

where h = height

u = initial velocity (which is 0 here)

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

t = time.
• Oct 22nd 2008, 05:35 PM
jeepwheeler
Got it - thanks a lot! Oddly worded question, that was the tough part.
• Oct 22nd 2008, 05:36 PM
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.
• Oct 22nd 2008, 06:02 PM
Shyam
Quote:

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

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

$\displaystyle h=19.62\;m$

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

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

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

so, object touches ground before 3 seconds

Height fallen by object in 3rd second = 25 - 19.62 = 5.38 m
• Oct 22nd 2008, 06:17 PM
skeeter
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 ...

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

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

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

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