On the planet Xenophous a 1.00 m long pendulum on a clock has a period of 1.32 s. What is the free fall acceleration on Xenophous?

The period of the pendulum is given by the following equation

T = 2 π * sqrt (L/g)

Where g is the gravity (free fall acceleration)

L is the longitude of the pendulum

T is the period.

We find g ... > (T / 2 π) ^ 2 = L/g

g = L / (T / 2π) ^2 ... > g = 22.657 m/s^2