Explanation

(c) Let the initial velocity of balls $A,\ B\ and\ C$ are equal and its magnitude is $u.$ Since, the ball $A$ is projected with velocity $u$ in upward direction, so when it will come back to the projection point, its velocity remains same. So, the final velocity of ball $A,$ when it hits the ground is given as
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$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v_{A}^{2} = u^{2} + 2gh$
Hence, $\ \ \ \ \ \ \ \ \ \ v_{A} = \sqrt{u^{2} + 2gh}\ \ \ \ \ \ldots(i)$
And the final velocity of ball $B$ is
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v_{B} = \sqrt{u^{2} + 2gh}\ \ \ \ \ \ \ldots(ii)$
But the initial vertical velocity of the ball $C$ is zero.
So, $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v_{C}^{2} = \sqrt{(0)^{2} + 2gh}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v_{C} = \sqrt{2gh}$ ….(iii)
Hence, it is clear from Eqs. (i), (ii) and (iii), we get
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v_{A} = v_{B} > v_{C}$