Math problem solving

Question

If the acceleration due to gravity is 9.8m/s² and the period equals 1s, find the length to the nearest cm (100cm= 1m)

Answer 1

s(t) = -4.9(t^2)

s(1) = -4.9m = -4.9(100 cm) = -490 cm

or just 490 cm as distances are often treated as positive, although a negative number may indicate direction

this assumes no initial velocity or distance, other than both zero

general formula is

s(t) =(a/2)t^2 +vot + so where vo=initial velocity and so = initial distance,

and a=-9.8 meters per second per second acceleration due to gravity