velocity of falling object

**Greg E** · December 21st 2005, 03:23 AM

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

**CaptainBlack** · December 21st 2005, 12:45 PM

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 $-g$ the minus sign is conventional as
up is normally taken as positive, and so downward accelleration is negative.

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

Integrating again gives:

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

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

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

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

or:

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

Putting this back into the equation for vertical velocity:

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

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

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

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

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

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

RonL

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