Answer:
1.23 m
Explanation:
The vertical distance covered by a free-falling object starting from rest in a time t is
[tex]y=\frac{1}{2}gt^2[/tex]
where
g = 9.8 m/s^2 is the acceleration due to gravity
In this problem, we have
t = 0.50 s
So the distance covered is
[tex]y=\frac{1}{2}(9.8 m/s^2)(0.50 s)^2=1.23 m[/tex]
