Question

a body dropped from rest falls half its total path in the last second before it strikes ground. From what height was it dropped?

Answer 1

The body falls under acceleration due to gravity which is constant and it's value is 9.8m/s^2 or 10m/s^2.
So for constant acceleration
S = ut + 1/2at^2
S = X/2m (the total distance is X , so half the distance is X/2 )
U=0 ( it is dropped from rest )
a = 10m/s^2
t = 1sec
By substituting the values
X/2= 1/2× 10 × 1^2
X = 10 m. ( It is the total distance)
So it is dropped from a height of 10m