Answer
The answer to part (d) will be greater than the answer to part (b)
$t = \frac{\sqrt{v_o^2+2gh}+v_0}{g}$
Work Step by Step
In part (b), we found the time to reach the ground is $t = \frac{\sqrt{v_o^2+2gh}-v_0}{g}$
When the ball is thrown upward, the ball needs to go up to maximum height and then fall back to the starting point. The remaining part of the motion takes the same time as the answer to part (b).
Therefore, the answer to part (d) will be greater than the answer to part (b)
We can find the time it takes to reach maximum height:
$v_f = v_0+at$
$t = \frac{v_f - v_0}{a}$
$t = \frac{0 - v_0}{-g}$
$t = \frac{v_0}{g}$
The total time to reach maximum height and return to the starting point is $\frac{2v_0}{g}$
We can find the total time to reach maximum height and then fall to the ground:
$t = \frac{2v_0}{g}+\frac{\sqrt{v_o^2+2gh}-v_0}{g}$
$t = \frac{\sqrt{v_o^2+2gh}+v_0}{g}$
