![speed:\text{ 231}(ft)/(s)](https://img.qammunity.org/2023/formulas/mathematics/high-school/uwq31y76m44u5gkh2vvyg8ogn0qulogsu9.png)
![Distance\text{ fallen in 10 seconds: 2310 ft}](https://img.qammunity.org/2023/formulas/mathematics/high-school/gk6i31rdiad31575bw6w4kzngm7d05t8hs.png)
Step-by-step explanation
to solve this we need to know the equivalence, in this case it is
![1\text{ m=3.3 ft}](https://img.qammunity.org/2023/formulas/mathematics/high-school/nh5u46ul8afa9cbwwz0gcaf8ym235spkk4.png)
then, we can make a equivalent fraction, this fraction needs to have the unit to convert in the denominator, for this problem the unit to convert is the meter, so
![\begin{gathered} \frac{3.3\text{ ft}}{1\text{ m}}\Rightarrow equivalent\text{ fraction} \\ note\text{ that } \\ \frac{3.3\text{ft}}{1\text{m}}=1,\text{ so the amount is not changed} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/it56mow10m8vcupy84m1ka39yww6r0div9.png)
hence
Step 1
convert from yards per second into ft per second, to do that, just multiply by the equivalent fraction
so
![\begin{gathered} 70(m)/(s) \\ 70(m)/(s)*\frac{3.3ft}{1\text{ m}}=231(ft)/(s) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/lqifulxhx6m184lr10vgnh6m5kwp9r8tox.png)
so, the speed is
![speed:\text{ 231}(ft)/(s)](https://img.qammunity.org/2023/formulas/mathematics/high-school/uwq31y76m44u5gkh2vvyg8ogn0qulogsu9.png)
Step 2
now, to find the distance we need to mutltiply the speed for the time,so
![distance=\text{ speed *time}](https://img.qammunity.org/2023/formulas/mathematics/high-school/ge60zk65ft5lm8cyctkp2c6sjbaarb7xa8.png)
replace
![\begin{gathered} distance=231(ft)/(s)*10\text{ s} \\ distance=2310\text{ ft} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/8q4cca0jqlw1ln7ctxs1dq57gtoqqppzbz.png)
so
![Distance\text{ fallen in 10 seconds: 2310 ft}](https://img.qammunity.org/2023/formulas/mathematics/high-school/gk6i31rdiad31575bw6w4kzngm7d05t8hs.png)
I hope this helps you