The velocity time graph for the veticaly component of the velocity of a body thrown upwards from the ground and landing on the roof of a building is given in the figure.The height of the building is:

The velocity time graph for the veticaly component of the velocity of

1.	The velocity time graph for the veticaly component of the velocity of a body thrown upwards from the ground and landing on the roof of a building is given in the figure.The height of the building is:
Answer» 50 m 40 m 20 m 30 m Solution :`u=30 m//s` `V=0m//s` `t=3s "" S=h=?` `v=u+a t` `0= 30+a(3)` `a=(-30)/(3)=-10m//s^(2)` `V^(2)=u^(2)+2aS` `O=(30)^(2) +2(-10)h` `(-30)^(2)=-20h` `h=(-30xx 3cancel0)/(-2cancel0)=45m` From MAX. height to roof of building `h=(u^(2))/(2g)=(10^(2))/(2xx10)=5m` height of building `=H-h=45-5` `= 40 m`