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A BALLOON IS ASCENDING WITH AN ACC. OF 0.2m/s^2. TWO STONES ARE DROPPED FROM IT AT AN INTERVAL OF 2s.FIND THE DISTANCE BETWEEN THEM 1.5s AFTER THE SECOND STONE IS RELEASED.
A)49m B)48m C)50m D)47m
Dear vikas
let when first stone is drop speed of the baloon is u
so after 2 sec
h=2u + 1/2 * .2 * 4
=2u+.4
and after 2 sec apeed of the baloon
v=u+2*.2
=u+.4
let h1 distance travell by first stone in (2+1.5) sec after it was dropped
h1 =u(3.5 ) -1/2 *g *(3.5)2
let h2 distance travell by second stone in 1.5 sec
so h2= (u+.4)*2 -1/2 *g*(1.5)2
so different between two stone =|h1-h2-h|
=50 m
All the best. Regards,Askiitians ExpertsBadiuddin
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