Answer
$E_{tot}=E_{k1}+E_{p01}=\dfrac {mv^{2}_{1}}{2}+mgh=mgH\Rightarrow v_{1}=\sqrt {2g\left( H-h\right) }\approx 12,91\dfrac {m}{s}$
Work Step by Step
The total Energy of the book at the top of the building is: $E_{tot}=E_{k0}+E_{p0}=0+mgH(1)$ since book doesn't have initial speed So when student catches the book the total energy of the book will be: $E_{tot}=E_{k1}+E_{p1}=\dfrac {mv^{2}_{1}}{2}+mgh\left( 2\right) $ where H is the height of the building and h is the height relative to the ground that student catches the book so from (1) and (2) we get: $E_{tot}=E_{k1}+E_{p01}=\dfrac {mv^{2}_{1}}{2}+mgh=mgH\Rightarrow v_{1}=\sqrt {2g\left( H-h\right) }\approx 12,91\dfrac {m}{s}$
