A ball thrown up is caught back by the thrower after 6 sec (1) the velocity with which the ball was thrown up (2) maximum height (3) the distance of ball below the highest point after 2 sec
Answers
Answered by
3
Time taken to reach highest would be 3 secs.
V=0 at max height
v=u-gt
u=gt
=9.8 x 3=29.4m/sec
v²=u²-2gh
2gh=u²
h=(29.4)²/19.6=44.1
V=0 at max height
v=u-gt
u=gt
=9.8 x 3=29.4m/sec
v²=u²-2gh
2gh=u²
h=(29.4)²/19.6=44.1
dharaj22pch5o4:
I am still thinking the 3 rd
Plz rate brainliest
Take g =10m/s square sorry i forgot to write
Then just exchange value of g to 10 m/s²
u will be then 30 m/sec
then ans is 45
height
You haven't find the ans of 3 question so please find ans of 3 ques its very urgent I have tried to find the ans of 3 ques my ans is 5 m so is it correct
Similar questions