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# A Problem from Projectile motion

“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?”

ans : 1.356 sec and 1.444 sec .
sol:
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)

## 1 Comment

1. vimal raj says:

ans : 1.356 sec and 1.444 sec .
sol:
from v=u+at we get t=2.8 . (time taken to rach 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 fm S=ut + 1/2 at2 ,we get t=1.356 .
and nxt time can be calculated by 2.8 – 1.356 =1.444 sec .

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