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 Experts
Badiuddin