Question

Value of acceleration due to gravity near to earth's surface is $9.81m{s}^{-2}$. Determine its value in $ft{s}^{-2}$. $(1ft=12inch,1inch=2.54cm)$

Answer 1

Step 1: Given data
Acceleration due to gravity, $g=9.81m{s}^{-2}$
$$\begin{gathered} 1ft=12inch\Rightarrow 1inch=\frac{1}{12}ft \\ 1inch=2.54cm\Rightarrow 1cm=\frac{1}{2.54}inch \end{gathered}$$
Step 2: Calculate the value of $1m$ in terms of $ft$.
$$\begin{gathered} 1m=100cm \\ =100\times \frac{1}{2.54}inch\left(As1cm=\frac{1}{2.54}inch\right) \\ =100\times \frac{1}{2.54}\times \frac{1}{12}ft\left(As1inch=\frac{1}{12}ft\right) \\ =3.28ft \end{gathered}$$
Step 3: Convert the value of acceleration due to gravity in terms of $ft{s}^{-1}$.
$$\begin{gathered} g=9.81m{s}^{-2} \\ =9.81\times 3.28ft{s}^{-2} \\ =32.18ft{s}^{-2} \end{gathered}$$
Thus, the value of acceleration due to gravity near to earth's surface in $ft{s}^{-2}$ is $32.18ft{s}^{-2}$.