Question

A ball thrown up is caught back by the thrower after 6 seconds.1.calculate the velocity with the ball was thrown up.2.the maximum height attain by the ball.3.the distance of the ball below the highest point after two seconds .

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