A stone falls from a balloon that is descending at a uniform rate of 12m/s. the displacement from the stone from the point of release after 10 sec is (a) 490 m (b) 510m (c) 610 m (d) 725 m

							
The point of release is a point in the air, not attached to the balloon.
Since speed of balloon has same direction as acceleration due to gravity,
u = 12m/s    a=9.8 m2/s  t=10s
s = ut  +  0.5at2 
  = 120   +   (0.5)(9.8)(100)
  = 610 m
 
						

							
since the stone is falls from a stone which is decending with 12m/s
 that means unitial velocity of stone is given 12m/s .......
for constant accleration we have , s(displacement) = ut + at2/2
a = -g = -9.8m/s2(downward)      ,  u = -12m/s (downward)
             s = -12t - 9.8t2/2
after 10 sec
          s = -120-490=610m
option c is correct
 
						

							
610 m
						

							Since balloon is descending at rate of 12m/s therefore the initial velocity of stone will be equal to 12m/sBy distance formulaS=ut+½at²=12*10+½*9.8*81=610
						

							
we know that s=ut+1/2at2 
here, u=12m/s; a=9.8m/s2; t=10 s
by substituting the values, we get
s=12×10+1/2(9.8)(10)2
s=610 m
Therefore,the displacement of the stone from the point of release after 10 seconds is 610 metres.