Answer
The minimum initial speed the puck should have is $2.4m/s$.
Work Step by Step
According to the work-energy theorem, $$E_f-E_0=W_{nc}$$
The puck never leaves the ground, so in this case, $PE=0$. Therefore, $E=KE$ $$\frac{1}{2}mv_f^2-\frac{1}{2}mv_0^2=W_{f}$$ $$\frac{1}{2}m(v_f^2-v_0^2)=W_{f}$$
In the first case, the hockey player gives the puck $v_0=1.7m/s$ and after traveling only half a distance, $v_f=0$, so $$W_{f, 1}=-1.445m $$ $$-fs_1=-1.445m$$ $$fs_1=1.445m (1)$$ (the negative sign shows the force negates the motion)
In the second case, the hockey player gives the puck $v_0$ enough to travel to a teammate and come to a halt $v_f=0$, so $$W_{f, 2}=-\frac{1}{2}mv_0^2 $$ $$-fs_2=-\frac{1}{2}mv_0^2 $$ $$fs_2=\frac{1}{2}mv_0^2 (2)$$
Divide (2) over (1), $$\frac{s_2}{s_1}=\frac{1/2v_0^2}{1.445}=\frac{v_0^2}{2.89}$$
We know $s_2=2s_1$, so $v_0=2.4m/s$
