The correct option is B 20 ms−1
Given Height h=20 m
Acceleration due to gravity g=10 ms−2.
Initially the body is dropped, so initial velocity u=0 ms−1.
According to the conservation of energy, energy will remain constant throughout. So, potential energy at top is equal to kinetic energy at bottom
mgh=12mv2
or v2=2gh=2×10×20
or v2=400
or v=20 ms−1