Question

a ball is thrown up and caught by the thrower after 6 sec.calculate 1)velocity with which the ball was thrown up 2) the maximum height attained by the ball 3) the distance of the ball below the highest point after 2 sec.take 10m/s square

Answer 1

using the formula h=u²/2g
and t=u/g
so u= 60 m/s
and therefore h=180 m
at highest point u=0
s=180+gt²/2
180+20
200m
check the last if wrong contact