Question 821517
If we're talking about an object dropping (with no initial velocity and no other forces than gravity) and if we're on Earth, then
{{{d = 9.8t^2}}}
where d is the distance traveled by the object, the 9.8 is the acceleration due to gravity (in meters/second/second). Since we know that the object dropped 54 meters:
{{{54 = 9.8t^2}}}
Divide by 9.8:
{{{5.51 = t^2}}} (rounded to two places)
Square root of each side (ignoring the negative square root since time must not be negative):
{{{2.35 = t}}} (rounded to two places)