Menu Close

A-balloon-is-ascending-vertically-with-an-acceleration-of-0-2-ms-2-Two-stones-are-dropped-from-it-at-an-interval-of-2-s-The-distance-between-them-when-the-second-stone-dropped-is-take-g-9-8-m




Question Number 19774 by Tinkutara last updated on 15/Aug/17
A balloon is ascending vertically with  an acceleration of 0.2 ms^(−2) . Two stones  are dropped from it at an interval of 2 s.  The distance between them when the  second stone dropped is (take g = 9.8  ms^(−2) )
Aballoonisascendingverticallywithanaccelerationof0.2ms2.Twostonesaredroppedfromitatanintervalof2s.Thedistancebetweenthemwhenthesecondstonedroppedis(takeg=9.8ms2)
Answered by ajfour last updated on 15/Aug/17
For first stone:  y_1 =h+ut−(1/2)gt^2       where h is height of balloon at t=0  u is initial velocity of balloon at t=0  For second stone:  y_2 =ut+(1/2)at^2   so y_2 −y_1 =(1/2)(g+a)t^2      y_(rel) =((1/2)×10×4)m =20m .
Forfirststone:y1=h+ut12gt2wherehisheightofballoonatt=0uisinitialvelocityofballoonatt=0Forsecondstone:y2=ut+12at2soy2y1=12(g+a)t2yrel=(12×10×4)m=20m.
Commented by Tinkutara last updated on 15/Aug/17
Thank you very much Sir!
ThankyouverymuchSir!

Leave a Reply

Your email address will not be published. Required fields are marked *