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vikas sharma Grade: 12
        

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

7 years ago

Answers : (1)

Badiuddin askIITians.ismu Expert
147 Points
										

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





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Badiuddin

7 years ago
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