Chapter 4 - Two-Dimensional Kinematics - Problems and Conceptual Exercises - Page 106: 31

The solution is $$v_0=1.27\text{ m/s}.$$

We know that $u_x=\frac{x}{t}$ $u_x=\frac{1.3m}{1.25s}$ $u_x=1.04m/s$ Now $u=\frac{u_x}{cos\theta}$ We plug in the known values to obtain: $u=\frac{1.04m/s}{cos35^{\circ}}$ $u=1.27m/s$