Explanation

(c) Let the ball starts moving with velocity ‘$u$’ and it reaches up to maximum height $H_{\max}$, then
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From $H_{\max} = \frac{u^{2}}{2g}$
$u = \sqrt{2g\ (H_{\max}})$
$= \sqrt{2 \times 10 \times 2} = 2\sqrt{10}\ m/s$
This velocity is supplied to the ball by the hand and initially the hand was at rest,
it acquires this velocity in distance of $0.2\ meter$
$\therefore a = \frac{u^{2}}{2s} = \frac{40}{2 \times 0.2} = 100\ m/s^{2}$
So upward force on the ball $F = m(g + a)$
$= 0.2(10 + 100) = 0.2 \times 110 = 22N$