A balloon is ascending at the rate of 9.8m/s. At height of 39.2m above the ground. When a foot packet is dropped from balloon. After how much time and with what velocity does it reach the ground. ( $g = 9.8 m / s^{2}$ )

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Solution

Given,
Height, $s = 39.2 m$ ,
Initial velocity, $u = - 9.8 m / s$ (as balloon was ascending with same speed but packet direction is now opposite)
$a = 9.8 m / s^{2}$
Using the second equation of motion we get,
$s = u t + \frac{1}{2} a t^{2}$
$39.2 = - 9.8 t + (1 / 2) 9.8 t^{2}$
$39.2 = 9.8 (- t + (t^{2} / 2))$
$4 = - t + (t^{2} / 2)$
$0 = t^{2} - 2 t - 8$
$(t - 4) (t + 2) = 0$
$t = - 2$ and $t = 4$
$t = - 2$ is not possible as time cannot be negative
so, $t = 4 s$
Now, using the first equation of motion we get,
$v = u + a t$
$= - 9.8 + 9.8 \times 4$
$= 29.4 m / s$ .

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