Chapter 8 - Potential Energy and Conservation of Energy - Problems - Page 208: 68b

$v_{_y}= 52.4m/s$

The initial horizontal velocity is equal to the final horizontal velocity, so $v_x= 53.78m/s$ $K=\frac{1}{2}m(v_{_x}^2 +v_{_y}^2)$ $(v_{_x}^2 +v_{_y}^2)=\frac{2K}{m}$ $v_{_y}^2=\frac{2K}{m}-v_{_x}^2 $ $v_{_y}=\sqrt {\frac{2K}{m}-v_{_x}^2} $ $v_{_y}=\sqrt {\frac{2\times1550J}{0.55kg}-53.78_{m/s}^2} $ $v_{_y}= 52.4m/s$