A ball B of mass $m$ is lying at rest on the top surface of a smooth horizontal table 5 m high. Another moving ball A of same mass makes on elastic collision with B and B slides off the table and strikes the floor at a horizontal distance of 10 m from the table. Then select the correct alternative (s)  $(g=10m{s}^{-2})$

Horizontal velocity of the ball B at the time when it strikes the ground is $v_u,$ then  $10=v_u\times t$
 $$\begin{gathered} \Rightarrow 10=v_B\times \sqrt{\frac{2\times 5}{g}}=v_u\sqrt{\frac{10}{10}}=v_b \\ \Rightarrow v_b=10m{s}^{-1} \end{gathered}$$
Since collision is perfectly elastic and mass of ball A and that of B are same, therefore, velocity of ball A, before collision is also $10m{s}^{-1}$ 

So, option (C) is correct
$$\begin{gathered} KE=\frac{1}{2}m{v}^2=\frac{1}{2}m{\left[\sqrt{{\left(10\right)}^2+{\left(10\right)}^2}\right]}^2 \\ \Rightarrow KE=\frac{1}{2}m\times 2\times 100=(m\times 100)J \end{gathered}$$