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!
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
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
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.