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:

Options

A. 50 m
B. 40 m
C. 20 m
D. 30 m

Answer

B

Explanation

`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`