# If an object is dropped, how fast will it be moving after falling 16 m?

Jan 12, 2017

The velocity is $= 17.7 m {s}^{-} 1$

#### Explanation:

Computing in the vertical direction ${\downarrow}_{+}$

$u = 0$

$a = g = 9.8$

$s = 16$

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

${v}^{2} = 0 + 2 \cdot 9.8 \cdot 16$

$v = \sqrt{2 \cdot 9.8 \cdot 16} = 17.7 m {s}^{-} 1$

Jan 12, 2017

$v = \frac{28 \sqrt{10}}{5} m {s}^{- 1}$

#### Explanation:

measuring down as positive

Considering vertical motion

initial velocity $u = 0 m {s}^{- 1}$

acceleration $g = + 9.8 m {s}^{_ 2}$

displacement $s = 16 m$

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

${v}^{2} = 0 + 2 \times 9.8 \times 16$

${v}^{2} = 313.6$

$v = \sqrt{313.6} = \frac{28 \sqrt{10}}{5} m {s}^{- 1}$