#### Answer

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

#### Work Step by Step

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, $v_{0}=41m/s$