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From a balloon moving upwards with a velocity of $12 ms^{-1}$, a packet is released when it is at a height of 65 m from the ground. Time taken by it to reach the ground is $(g = 10 ms^{-2})$
Explanation
$s = ut + \frac{1}{2}a t^{2}$ $-65 = 12t - 5t^{2}$ on solving we get, $t = 5 s$