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
Answers
Answered by
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:
Similar questions