At maximum height, velocity v = 0 m/s.

(i) Time taken to reach the maximum height

Step 1: At maximum height, velocity $v = 0$ m/s.
Equation of motion:
$v = u - g t$
Substitute $v = 0$:
$0 = u - g t$
u = g t \tag{1}

Step 2: Displacement $h = 50$ m.
Equation:
$h = u t - \frac{1}{2} g t^{2}$
Substitute $u = g t$ from (1):
$50 = (g t) t - \frac{1}{2} g t^{2}$
$50 = g t^{2} - \frac{1}{2} g t^{2}$
$50 = \frac{1}{2} g t^{2}$

Step 3: Substitute $g = 10$ m/s²:
$50 = \frac{1}{2} \times 10 \times t^{2}$
$50 = 5 t^{2}$
$t^{2} = 10$
$t = \sqrt{10} \ s$

\sqrt{10 \ s}

(ii) Speed of the throw

From equation (1): $u = g t$
Substitute $g = 10$ m/s² and $t = \sqrt{10}$ s:
$u = 10 \times \sqrt{10} = 10\sqrt{10} \ m/s$

Alternatively, using $v^{2} = u^{2} - 2 g h$:
$0 = u^{2} - 2 \times 10 \times 50$
$u^{2} = 1000$
$u = \sqrt{1000} = \sqrt{100 \times 10} = 10\sqrt{10} \ m/s$

10\sqrt{10 \ m/s}