Question

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Answer 1

Answer:

$g=16.31\ m/s^2$

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 :

$T=2\pi \sqrt{\dfrac{l}{g}}$

or

$T^2=4\pi ^2\dfrac{l}{g}\\\\g=\dfrac{4\pi ^2l}{T^2}\\\\g=\dfrac{4\pi ^2\times 2}{(2.2)^2}\\\\g=16.31\ m/s^2$

So, the free fall acceleration is $16.31\ m/s^2$.