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A balloon is starting from the ground has been ascending vertically at uniform velocity for 4 sec and a stone let to fall from it reaches the ground in 6 sec. Find the velocity with which the stone hits the ground. ($g= 10 \frac{m}{s^2}$)

The answer given in my book is $42 \frac{m}{s}$ but I got $-60 \frac{m}{s}$ using the equation $v = u + at$. If my answer is wrong please tell why this equation didn't work.

Working :-
$t = 6 s$
$u = 0 \frac{m}{s}$ (left fall means stone is released from rest without any push)
It's under free fall $a = -10\frac{m}{s^2}$
$v = u + at$
$v = 0 + (-10)6$
$v = -60 \frac{m}{s}$

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    $\begingroup$ could you please show your complete working? $\endgroup$
    – user60063
    Oct 4, 2014 at 12:23
  • $\begingroup$ Working added in question. $\endgroup$
    – pcforgeek
    Oct 4, 2014 at 12:46
  • $\begingroup$ What is the 'uniform velocity' with which the balloon is moving? $\endgroup$
    – Gaurav
    Oct 4, 2014 at 13:19
  • $\begingroup$ We have to assume it as u and we get initial velocity of of stone w.r.t ground which is equal to u. Then h = 4u. Substituting these value in 2nd equation of motion with constant acceleration we get u = 18 m/s. We get v = -42 m/s from v = u + at. $\endgroup$
    – pcforgeek
    Oct 4, 2014 at 13:55

1 Answer 1

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Because the initial velocity of the stone is not zero.

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  • $\begingroup$ The phrase "left fall" means the stone is released from rest without any push $\endgroup$
    – pcforgeek
    Oct 4, 2014 at 12:47
  • $\begingroup$ @pcforgeek: I believe it was "released from rest without any push" with respect to the balloon, not with respect to the ground. $\endgroup$
    – akhmeteli
    Oct 4, 2014 at 12:49
  • $\begingroup$ @pcforgeek: At least if you assume that, you get the answer from your book. $\endgroup$
    – akhmeteli
    Oct 4, 2014 at 12:50
  • $\begingroup$ You might be right as if we use 2 nd equation of motion in straight line with uniform acceleration then this answer should be right as we will get u = 18 m/s if u is initial velocity of stone relative to ground. $\endgroup$
    – pcforgeek
    Oct 4, 2014 at 12:55

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