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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 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
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
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 sby substituting the values, we gets=12×10+1/2(9.8)(10)2s=610 mTherefore,the displacement of the stone from the point of release after 10 seconds is 610 metres.
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