Answer
1.73 s
Work Step by Step
Since the ballast bag is released from rest relative to the balloon, its initial velocity relative to the ground is equal to the velocity of the balloon relative to the ground.
Let's apply equation 3.5b $S=ut+\frac{1}{2}at^{2}$ in the vertical direction to find the time required for the ballast to reach the ground.
$\uparrow S=ut+\frac{1}{2}at^{2}$ ; Let's plug known values into this equation.
$-9.5\space m=3\space m/s\times t+\frac{1}{2}(-9.8\space m/s^{2})\times t^{2}$
$4.9t^{2}-3t-9.5=0$
This is a quadratic equation. We can find t using the quadratic formula.
$t=\frac{-(-3)\space \pm \sqrt {(-3)^{2}-4(4.9)(-9.5)}}{2(4.9)}=\frac{3\space \pm14}{9.8}$
Since time is a positive value, here we neglect the negative solution for t.
So, $t=\frac{3+14.}{9.8}s=1.73\space s$
