We will apply the concept of period in a pendulum, defined as the product between 2 by the square root of the length over gravity, this is mathematically
Here,
T = Period
L = Length
g = Acceleration due to gravity
For the period to be 1 second, then we must look for the necessary length for such a requirement so
The meter's length would be slight less than one-fourth of its current length. Also, the number of significant digits depends only on how precisely we know g, because the time has been defined to be exactly 1s.
Therefore the correct answer is C.
Seriously this is fun dont know why u want me to enjoy this so ummmm hmmm lolololol the equation for this would be s=vif+1/2at^2 so time is lets see plug it and vell vi=0 since it started from rest so s=1/2at^2 and plug it in the displacement and time 110=1/2 a 5.21^2 and solve it and u get 8.1m/s^2