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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

6 years ago

jishnu prakash
15 Points

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

6 years ago
510 Points

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

6 years ago
divyanshu
21 Points

610 m
2 years ago
Anubhav
13 Points
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

4 months ago
swetlana
26 Points

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.
4 months ago
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