a balloon is ascending vertically with an acceleration of 0.2 m/s2. Two stones are dropped from it at an interval of 2sec. Find the distance between them 1.5 sec after the second stone is released. (use g=9.8)

Question

gauhar singh · Answer

first u have to specify the velocity of balloon when stone is dropped,let it be v,

velocity of 1st stone is v in upward direction n acceleration downward, use

vt - 0.5 gt²= d using appropriate signs for directions. t=2+1.5 sec

the 2nd stone will have addition separation due to movement of balloon n also additional speed

use same formula with t=1.5 sec and v₂= v + time.acceleration of balloon

*make a diagram with reference point as the place where 1st stone is drawn n measure distances accordingly

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a balloon is ascending vertically with an acceleration of 0.2 m/s2. Two stones are dropped from it at an interval of 2sec. Find the distance between them 1.5 sec after the second stone is released. (use g=9.8)

a balloon is ascending vertically with an acceleration of 0.2 m/s2. Two stones are dropped from it at an interval of 2sec. Find the distance between them 1.5 sec after the second stone is released. (use g=9.8)

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