Therefore, by using the law of conservation of energy at this moment, we can write
$\operatorname{mg}(h+x)=1 / 2 m v^{2}+1 / 2 k x^{2}$
$\Rightarrow 1 / 2 mv ^{2}= mg ( h + x )-1 / 2 kx ^{2} \ldots$ (i)
[The velocity of the block will be maximum when the kinetic energy is maximum. Also, the velocity keeps on increasing as long its acceleration is increasing]
$\therefore d \left[1 / 2 mv ^{2}\right] / d x = mg - k x =0$
$\Rightarrow mg - kx =0$
$\Rightarrow x=m g / k$
Now, to get the height "h" from the surface of the table at which the velocity is maximum, we need to subtract the distance of compression in the spring from the length of the spring, so,
$h = I - mg / k \ldots \ldots .$ [from (i) $]$
$\Rightarrow h=20-\left[1^{*} 9.8 / 2\right] \ldots \ldots\left[\operatorname{taking} g=9.8 m / s ^{2}\right]$
$\Rightarrow h=20-4.9$
$\Rightarrow h =15.1 cm =15 cm$
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