# If an object is dropped, how fast will it be moving after falling 21 m?

Jun 29, 2018

$\text{20.3 m/s}$

#### Explanation:

Use equation of motion

$\text{v"^2 - "u"^2 = "2aS}$

Where

• $\text{v =}$ Final velocity
• $\text{u =}$ Initial velocity
• $\text{a =}$ Acceleration
• $\text{S =}$ Displacement

The object is dropped. So $\text{u}$ is zero.

"v" = sqrt(2"aS")

$\textcolor{w h i t e}{\text{v") = sqrt(2 × "9.8 m/s"^2 × "21 m}}$

color(white)("v") = 20.3\ "m/s"

Jun 29, 2018

${\text{20.3 m s}}^{-} 1$ (to 3 significant figures)

#### Explanation:

• Using the formula:

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

where;

• $v$ is the final velocity of the object which we need to find.
• $u$ is the initial velocity of the object, which is $0$ since it fell from rest.
• $a$ is the acceleration of free fall $\left({\text{9.81 m s}}^{-} 2\right)$
• $s$ is the displacement, which is $\text{21 m}$

Now all you gotta do is substitute:

${v}^{2} = {0}^{2} + \left(2 \times \text{9.81 m s"^(-2)xx"21 m}\right)$

${v}^{2} = {\text{412.02 m"^2"s}}^{- 2}$

So

$v = {\text{20.3 m s}}^{-} 1$ (to 3 s.f.)