# If an object is dropped, how fast will it be moving after falling 25 m?

Feb 4, 2016

Assuming no air resistance, it will be moving at $v = \sqrt{2 a d} = \sqrt{2 \cdot 9.8 \cdot 25} = \sqrt{490} = 22.1$ $m {s}^{-} 1$

#### Explanation:

The expression for how fast an object will be travelling after covering a certain distance while accelerating is:

${v}^{2} = {u}^{2} + 2 a d$

Where:

$v$ = final velocity $\left(m {s}^{-} 1\right)$
$u$ = initial velocity $\left(m {s}^{-} 1\right)$
$a$ = acceleration $\left(m {s}^{-} 2\right)$
$d$ = distance $\left(m\right)$

In this case, we assume that the initial velocity is $0$: the question says the object was 'dropped', not 'thrown'.

We can leave out the $u$ term and take the square root of both sides to make $v$ the subject:

$v = \sqrt{2 a d}$

Substituting in our values, including the acceleration due to gravity:

$v = \sqrt{2 \cdot 9.8 \cdot 25} = \sqrt{490} = 22.1$ $m {s}^{-} 1$

We have assumed that there is no air resistance acting, or that we can neglect it because it is much smaller than the gravitational force.