Arugby player passes the ball 7.75 m across the field, where it is caught at the same height as it left his hand. at what angle was the ball thrown if its initial speed was 12.0 m/s, assuming that the smaller of the two possible angles was used?

Given that the "launch" and "impact" heights are the same, we shall use the range equation and the time of flight equation.
(a)(b)range x=V²sin (2θ)/g
7.75m=(12 m/s)sin (2θ)/ 9.8m/s²
sin (2θ)=0.5274
hence:
2θ=arcsin 0.5274=31.83°
or 2θ=180-31.83=148.17°
hence the smaller angle will be:
31.83/2=15.915~15.92°
and the larger angle will be:
148.17/2=74.085°