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A balloon rises up with constant net acc...

A balloon rises up with constant net acceleration of `10 m//s^(2)`. After 2 s a particle drops from the balloon. After further 2 s match the following : `("Take" g=10 m//s^(2))`
`{:(,"Column I",,,"Column II"),((A),"Height of perticle from ground",,(p),"Zero"),((B),"Speed of particle",,(q),10 SI "units"),((C),"Displacement of Particle",,(r),40 SI "units"),((D),"Acceleration of particle",,(s),20 SI "units"):}`

Text Solution

Verified by Experts

The correct Answer is:
(A)-r, (B)-p, (C)-s, (D)-q`
Velocity & height of the bolloon after 2 sec:
`v=0+10xx2=20 m//s uarr h=1//2xx10xx4=20 m`
Initial velocity of drop particle is equal to the velocity of balloon `=20 m`
`:' u_(s)=20 m//s a_(s)=g darr`
After further `2s v_(s)=0`
`:.` height `=(u_(s)+v_(s))/2xx2=20 m` from initial position of balloon
`:.` Height from ground `=20+2v=40 m`
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