A body is projected up a smooth inclined plane of length `20sqrt(2)m` from point A as shown in the figure. The top of B of the inclined plane is connected to a well of diameter 40 m. If the body just manages to cross the well then the velocity of projection is
(Acceleration due to gravity, `g=10ms^(-2)`)

Options

A. `40ms^(-1)`
B. `40sqrt(2)ms^(-1)`
C. `20ms^(-1)`
D. `20sqrt(2)ms^(-1)`

Answer

D

Explanation

According to the question,

In `DeltaABC`,
`h/(20sqrt(2))=cos45^(@)`
`rArr h=20sqrt(2)(1/sqrt(2))` or h = 20 m
From third equation of the motion,
`v^(2)=u^(2)+2gh`
`rArr v=sqrt(2gh) " "(because u=0)`
`rArr v=sqrt(2 times 10 times 20)=20m//s`
Net energy at a point B,
`rArr 1/2mv^(2)+mgh=1/2m(20)^(2)+m times 10 times 20`
`" "=m(200+200)=400 m`
Net energy at point D to cross the well,
`E_(d)=1/2mv_(D)^(2)`
So, `" "1/2mv_(D)^(2)=400 m`
`rArr " "v_(D)=20sqrt(2)ms^(-1)`