Answer
$\approx 197.99$ m/s or, 198 m/s
Work Step by Step
Using the distance-time equation.
$s=ut+\dfrac{1}{2}gt^2$
Thus,
$s=s_0+(v_0 \sin \theta )t-\dfrac{1}{2}gt^2$
$500=0+v_0 \sin 30^\circ(v_0/2g)-\dfrac{1}{2}g(v_0/2g)^2$
$500=\dfrac{v_0}{2}\dfrac{v_0}{2g}-\dfrac{1}{2}g\dfrac{v_0}{2g}^2$
$500=\dfrac{v_0^2}{4g}-\dfrac{v_0^2}{8g}$
After simplifications, we get
$v_0=20\sqrt{10 \times 9.8}$ m/s
Hence, $v_0 \approx 197.99$ m/s or, 198 m/s
