“A ball is projected vertically upwards with an initial velocity of 28m/s and is observed to pass a point 30m above the projection point,at what 2times does the ball pass these points?”
(Vimal Raj Answered)
ans : 1.356 sec and 1.444 sec .
from v=u+at we get t=2.8 . (time taken to reach the max height )
fm v2-u2=2aS we get S=39.2 m (max height reached)
observer at 30 m .
so , in btw distance btw observer and ball (at the top) =9.2 m
at top ball is at rest and time taken to reach the observer can be calculaed from S=ut + 1/2 at2 ,we get t=1.356s .
and next time can be calculated by 2.8 – 1.356 =1.444 sec .
(The answer has not been scrutinized. If any error, teachers and visitors can post as comment)