A balloon is moving upwards with a speed of `20 m//s`. When it is at a height of 14 m from ground in front of a plane mirror in situation as shown in figure, a boy drops himself from the balloon. Find the time duration for which he will see the image of source S placed symmetrically before plane mirror during free fall.

Answer

A

Explanation

`(FG)/(BF)=(IH)/(BH) =(HS)/(BH)`
` :. FG=(BF)((HS)/(BH))`
`=(5)((0.1)/(0.5))`
`FC=2+10=12m`
The boy has dropped himself at point f. So, his velocity is `20m//s` in upward direction. Let us from the time to move from F to topmost point and then from topmost point to point
C. from `s=(ut)+(1/2)(at)^(2)`, we have
`=-12=(20t)+(1/2)(-10)t^(2)`
Solving this equation we get, `(t_(1))=4.53s.`
Velocity of boy at point G,
`v=sqrt((20^(2))-2xx10xx10)`
`=14.4m//s ( :. v^(2) =u^(2)-2gh)`
Time taken to move the boy from G to topmost point and then from topmost point to G will be
`(t_(2))=(2v)/(g)=2.83 s`
`:.` The required time is `t=t_(1)-t_(2)=1.7s`.