Here Intial velocity u=40 m/s
Final velocity ,u = 40 m/s
Final velocity v=0 (The stone stops )
Acceleration due to gravity,`g=-10 m//s ^(2)` (The stone stops )
And Height ,h= ? (To be calculated )
Now , `v^(2)=u^(2)+2gh`
So, `(0)^(2)=(40)^(2)+2 xx (-10)xxh`
0=1600 -20 h
20 h 1600
`h=(1600)/(20)`
h=80 m (ii) The stone is thrown up from the ground and after reaching the maximum height falls back to the ground .As the final position of stone coincides with the initial position of stone , the net displacement os the stone is zero (0)
(ii) The distance covered by the stone in reaching the maximum height is 80 meters .the stone will cover the same distance of 80 merters in coming back to ground .SO the total distance covered by the stone is 80 + 80=160
