Question

A ball thrown up is caught back by the thrower
after 6 sec Calculate;
(i) The velocity with which the ball was thrown up.
(ii) The maximum height attain by the wall
(iii) The distance of the ball below the heignes point
after 2 sec
take g = l0 m per sec​

Answer 1

Answer:

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

Explanation: