Answer:
[tex]g=16.31\ m/s^2[/tex]
Explanation:
Given that,
The length of the pendulum, l = 2 m
The period of the pendulum, T = 2.2 s
The formula for the time period of a pendulum is given by :
[tex]T=2\pi \sqrt{\dfrac{l}{g}}[/tex]
or
[tex]T^2=4\pi ^2\dfrac{l}{g}\\\\g=\dfrac{4\pi ^2l}{T^2}\\\\g=\dfrac{4\pi ^2\times 2}{(2.2)^2}\\\\g=16.31\ m/s^2[/tex]
So, the free fall acceleration is [tex]16.31\ m/s^2[/tex].
