The parable of the prodigal son is about God's love, and is found in the Bible in the book of .

At maximum height, $v = 0$, $u = 20$ m/s, $g = 10$ m/s$^2$.

$h = \frac{400}{20} = 20$ m

$\boxed{20 \text{ m}}$

(b)

At maximum height, $v = 0$.

$t = 2$ s

$\boxed{2 \text{ s}}$

(c)

$T = 2t = 2 \times 2 = 4$ s

$\boxed{4 \text{ s}}$

(d)

magnitude same as initial, direction downward. $v = -u = -20$ m/s (downward)

$\boxed{-20 \text{ m/s}}$

(a)

Velocity on ground.

$v = \sqrt{900} = 30$ m/s

$\boxed{30 \text{ m/s}}$

(b)

$t = 3$ s

$\boxed{3 \text{ s}}$

Find total distance.
First phase: free fall $t_1 = 2$ s, $u=0$, $g=10$ m/s$^2$.
$h_1 = \frac{1}{2}gt_1^2 = \frac{1}{2}(10)(2)^2 = 20$ m
$v_1 = gt_1 = 10 \times 2 = 20$ m/s

from $v_1 = 20$ m/s to $v_2 = 0$, $a = -2$ m/s$^2$. Time $t_2$.

$t_2 = 10$ s

$h_2 = v_1 t_2 + \frac{1}{2} a t_2^2 = 20(10) + \frac{1}{2}(-2)(10)^2 = 200 - 100 = 100$ m

$h = h_1 + h_2 = 20 + 100 = 120$ m

$\boxed{120 \text{ m}}$

Was ascending at $v_0 = 4$ m/s. Time to stop relative to ground.
Net acceleration now $a = g$ downward regardless. But since ascending, time to stop.

$t = 0.4$ s

$\boxed{0.4 \text{ s}}$

Distance in 1 min = 60 s.
First phase: $t_1 = 30$ s.
$v = at_1 = 3 \times 30 = 90$ m/s
$s_1 = \frac{1}{2} a t_1^2 = \frac{1}{2}(3)(30)^2 = 1350$ m

$t_2 = 30$ s, constant $v=90$ m/s. $s_2 = v t_2 = 90 \times 30 = 2700$ m

$s = s_1 + s_2 = 1350 + 2700 = 4050$ m

$\boxed{4050 \text{ m}}$