Question

a stone is released from top of the tower of height 19.6. calculate its velocity

Answer 1

v^2-0=2*196.08
v^2=384.16
v=19.6m/s

Answer 2

here height (distance)=19.6 m

acceleration due to gravity=9.8 m

initial velocity=0

we know v^2-u^2=2as

=>v^2-0=2*9.8*19.6

=>v^2=384.16

=>v=√384.16=19.6 m/s