Chapter 3 - Motion in Two Dimensions - Learning Path Questions and Exercises - Exercises - Page 100: 61

$v_{0}=41m/s$ at an angle of $\theta=11.9^{\circ}$ from horizontal.

Let the angle of direction of initial velocity with horizontal be $\theta$ $H=ut+\frac{1}{2}at^{2}$ $3=(v_{0}sin\theta) t-4.9t^{2}$ putting t=0.5s in the equation, $(v_{0}sin\theta)=8.45$ Also,$(v_{0}cos\theta)\times0.5=20$ $(v_{0}cos\theta)=40$ Therefore, $\theta=11.9^{\circ}$ So, $v_{0}=41m/s$