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A square platform of side length 8m is i...

A square platform of side length `8m` is is situated in `x-z` plane sucha that it is at `16 m` form the `x`-axis and `8 m` from the `z`-axis as shown in figure. A particle is projected with velocity `overset(vec)(v)=(v_(2)overset(hat)(i)+25hat(j)) m//s` relative to wind form origin and at the same instant the platform starts with acceleration `overset(vec)=(2hat(i)+2.5hat(j)) m//s^(2)`. Wind is blowing with velocity `v_(1)hat(k).g=10m//s^(2)`

`{:(,,"ListI",,"List II"),((P),,"Least possible values of " V_(2) "(in M/s) so that particle hits the platform or edge of platfrom is",,(1) 4),((Q),,"Least possible values of " v_(1) "(in m/s) so that particle hits the platform or edge or platform is ",,(2) 6),((Q),,"If t is the time (in seconds) after particle hits the platform then 2t is equal to ",,(3) 8),((S),,"Value of displacement with respect to ground (in m) of the particle in y-direction , when " v_(2) " has its minimum possible value is (till particle hits the platform or edge of platform)",,(4) 20):}`

A

`{:(,P,Q,R,S),((A),2,4,3,1):}`

B

`{:(,P,Q,R,S),((B),2,1,3,4):}`

C

`{:(,P,Q,R,S),((C),2,3,4,1):}`

D

`{:(,P,Q,R,S),((D),2,1,4,3):}`

Text Solution

Verified by Experts

`vecV_(P.P)=V_(2)hati+25hatj+V_(1)hatk`
`veca_(p.p)=-2hati-12.5hatj`
`vecV_(P.P)=` Velocity of particle relative to platform
Time `=(2xx25)/(12.5)=4` sec.
`8leV_(2)xx4-(1)/(4)xx2xx4^(2)le16`
`6leV_(2)le8`
`16leV_(1)xx4le24`
`4leV_(1)le6`
`Y=25xx4-(1)/(2)xx10xx4^(2)=100-80=20m`
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