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

(a)

Step 1: Maximum height formula for vertical throw upwards.
$v^2 = u^2 - 2gh$
At maximum height, $v = 0$, $u = 20$ m/s, $g = 10$ m/s$^2$.

Step 2: Substitute values.
$0 = (20)^2 - 2(10)h$
$0 = 400 - 20h$

Step 3: Simplify.
$20h = 400$
$h = \frac{400}{20} = 20$ m

$\boxed{20 m}$

(b)

Step 1: Time to maximum height formula.
$v = u - gt$
At maximum height, $v = 0$.

Step 2: Substitute values.
$0 = 20 - 10t$

Step 3: Simplify.
$10t = 20$
$t = 2$ s

$\boxed{2 s}$

(c)

Step 1: Total time to return to starting point is twice the time to max height (symmetric motion).
$T = 2t = 2 \times 2 = 4$ s

$\boxed{4 s}$

(d)

Step 1: Velocity on return: magnitude same as initial, direction downward.
$v = -u = -20$ m/s (downward)

$\boxed{-20 m/s}$

(a)

Step 1: Free fall from height $h = 45$ m, $u = 0$, $g = 10$ m/s$^2$. Velocity on ground.
$v^2 = u^2 + 2gh$

Step 2: Substitute values.
$v^2 = 0 + 2(10)(45)$
$v^2 = 900$

Step 3: Simplify.
$v = \sqrt{900} = 30$ m/s

$\boxed{30 m/s}$

(b)

Step 1: Time of fall formula.
$h = ut + \frac{1}{2}gt^2$

Step 2: Substitute values.
$45 = 0 + \frac{1}{2}(10)t^2$
$45 = 5t^2$

Step 3: Simplify.
$t^2 = 9$
$t = 3$ s

$\boxed{3 s}$

Step 1: Body falls freely for 2 s, then parachute opens, deceleration $a = 2$ m/s$^2$ until stop. 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

Step 2: Second phase: from $v_1 = 20$ m/s to $v_2 = 0$, $a = -2$ m/s$^2$. Time $t_2$.
$v_2 = v_1 + at_2$
$0 = 20 - 2t_2$
$t_2 = 10$ s

Step 3: Distance second phase.
$h_2 = v_1 t_2 + \frac{1}{2} a t_2^2 = 20(10) + \frac{1}{2}(-2)(10)^2 = 200 - 100 = 100$ m

Step 4: Total distance.
$h = h_1 + h_2 = 20 + 100 = 120$ m

$\boxed{120 m}$

Step 1: Elevator acceleration $a = 2$ m/s$^2$ upward, cable breaks, now $a = -10$ m/s$^2$. 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.
$v = v_0 + at$
$0 = 4 - 10t$
$t = 0.4$ s

$\boxed{0.4 s}$

Step 1: Train from rest, $a = 3$ m/s$^2$ for 30 s, then constant $v$. 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

Step 2: Second phase: $t_2 = 30$ s, constant $v=90$ m/s.
$s_2 = v t_2 = 90 \times 30 = 2700$ m

Step 3: Total distance.
$s = s_1 + s_2 = 1350 + 2700 = 4050$ m

$\boxed{4050 m}$