“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 .

sol:

from v=u+at we get t=2.8 . (time taken to reach the max height )

fm v^{2}-u^{2}=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 at^{2} ,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)**

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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 .