Question

the average velocity of a freely falling body is numerically equal to half of the acceleration due to gravity what is the velocity of the body as it reaches the ground

Answer 1

 A freely falling body has average velocity
V(av.) = (1/2). numerical value of g in m/s
suppose it travelled for t secs
so velocity of striking the ground v = u + g×t = g×t
if it has travelled a distance h in t seconds
h= (1/2) g×t^2
but the same distance it must have travelled with average velocity V(av.)
h = V(av.)× t =(g/2)×t = (1/2) g×t^2
therefore t= 1 s
so, v = g×1 m/s
the velocity with which it strikes the ground = g m/s




P.S :  IN SOME BOOKS, THE ANSWER IS WRITTEN AS "√2*g".........BUT I HAVE ASKED MY TEACHERS AND THEY SAY THAT "√2*g" IS A PRINTING MISTAKE AND THE CORRECT ANSWER IS "g"