Question

On the planet xenos, an astronaut observes that a 1.59 m long pendulum has a period of 1.57 s. what is the free-fall acceleration on xenos?

Answer 1

The period T of a pendulum of length L is given by
$T=2 \pi \sqrt{ \frac{L}{g} }$
where g is the free-fall acceleration value.
Given that on Xenos T=1.57 s and L=1.59 m, we can re-arrange the formula and replacing these numbers to find the value of g on Xenos:
$g=L \frac{(2 \pi)^2}{T^2}=(1.59m) \frac{(2 \pi)^2}{(1.57s)^2}=25.44 m/s^2$