Gravity on Another Planet

Gravity on Another Planet
Part A = 11.3m/s²
Solution Below:

After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 49.0 cm. The explorer finds that the pendulum completes 110 full swing cycles in a time of 144 s.

Part A

What is the magnitude of the gravitational acceleration on this planet?

The period of a pendulum is given by T = 2πsqrt(L/g). Since this is a different planet, g will be different, so we can use the above formula to solve for it. Start by calculating the period, T, the time it takes for one swing:
T = 144 s / 110 swings
T = 1.31
Now solve for g (L needs to be in meters, so divide 49.0 by 100):
T = 2πsqrt(L/g)
1.31 = 6.28sqrt(0.49/g)
0.2086 = sqrt(0.49/g)
0.0435 = 0.49/g
g = 0.49 / 0.0435
g = 11.3m/s²