Answer
The package takes 5.21 seconds to reach the ground.
Work Step by Step
$y = y_0 + v_0t + \frac{1}{2}at^2$
$0 = 105 + 5.40\cdot t -4.90 \cdot t^2$
We can use the quadratic formula to solve for t.
$t = \frac{-b ~\pm \sqrt{b^2 - 4ac}}{2a}$
$t = \frac{-5.40 ~\pm \sqrt{(5.40)^2 -(4)(-4.90)(105)}}{(2)(-4.90)}$
$t = 5.21 ~s, -4.11 ~s$
The negative value for t is an unphysical solution, so the correct value for t is 5.21 seconds. Therefore, the package takes 5.21 seconds to reach the ground.
