\({v_x} = \frac{{dx}}{{dt}} = 36\,m/s\)

\(y = 48t - 4.9{t^2}\)

\({v_y} = 48 - 9.8t\)

at \(t = 0\) \({v_x} = 36\) and \({v_y} = 48\,m/s\)

So, angle of projection \(\theta = {\tan ^{ - 1}}\left( {\frac{{{v_y}}}{{{v_x}}}} \right) = {\tan ^{ - 1}}\left( {\frac{4}{3}} \right)\)

Or \(\theta = {\sin ^{ - 1}}(4/5)\)