# velocity of falling object

• Dec 21st 2005, 03:23 AM
Greg E
velocity of falling object
If you drop a 1.36 kg weight and it falls .1524 meters, what is its velocity when it hits the ground and how is this calculated. Ultimately I am trying to decide how much energy is released.

Thanks,
Greg
• Dec 21st 2005, 12:45 PM
CaptainBlack
Quote:

Originally Posted by Greg E
If you drop a 1.36 kg weight and it falls .1524 meters, what is its velocity when it hits the ground and how is this calculated. Ultimately I am trying to decide how much energy is released.

Thanks,
Greg

An object dropped near the surface of the earth experiences an approximately
constant acceleration of $\displaystyle -g$ the minus sign is conventional as
up is normally taken as positive, and so downward accelleration is negative.

So $\displaystyle v=-g\cdot t+c$, and if the initial speed is $\displaystyle 0$ then $\displaystyle c=0$.

Integrating again gives:

$\displaystyle x=-\frac{g\cdot t^2}{2}+x_0$,

where $\displaystyle x_0$ is the initial height of the weight, for convenience we will
take this to be $\displaystyle 0$.

Now we are interested in what is happening when the weight has fallen
$\displaystyle 0.1524m$. This occurs when:

$\displaystyle 0.1524=-\frac{g\cdot t^2}{2}$,

or:

$\displaystyle t=\sqrt{\frac{2\cdot 0.1524}{g}}$.

Putting this back into the equation for vertical velocity:

$\displaystyle v=-g \cdot \left( \sqrt {\frac{2 \cdot 0.1524}{g}} \right)=-\sqrt{g \cdot 2 \cdot 0.1524}$.

At the earth surface $\displaystyle g \sim 9.81 m/s$.

We could have short circuited this calculation by knowing that the change
in potential energy when the weight drops through a height$\displaystyle h$is

$\displaystyle PE=m \cdot g \cdot h$,

which will equal the kinetic energy of the weight when it has fallen from rest
through a distance $\displaystyle h$:

$\displaystyle KE= \frac{1}{2}m \cdot v^2$

RonL