Take the missile's starting position to be the origin. Assuming the angles given are taken to be counterclockwise from the positive horizontal axis, the missile has position vector with components
[tex]x=v_0\cos20.0^\circ t+\dfrac12a_xt^2[/tex]
[tex]y=v_0\sin20.0^\circ t+\dfrac12a_yt^2[/tex]
The missile's final position after 9.20 s has to be a vector whose distance from the origin is 19,500 m and situated 32.0 deg relative the positive horizontal axis. This means the final position should have components
[tex]x_{9.20\,\mathrm s}=(19,500\,\mathrm m)\cos32.0^\circ[/tex]
[tex]y_{9.20\,\mathrm s}=(19,500\,\mathrm m)\sin32.0^\circ[/tex]
So we have enough information to solve for the components of the acceleration vector, [tex]a_x[/tex] and [tex]a_y[/tex]:
[tex]x_{9.20\,\mathrm s}=\left(1810\,\dfrac{\mathrm m}{\mathrm s}\right)\cos20.0^\circ(9.20\,\mathrm s)+\dfrac12a_x(9.20\,\mathrm s)^2\implies a_x=21.0\,\dfrac{\mathrm m}{\mathrm s^2}[/tex]
[tex]y_{9.20\,\mathrm s}=\left(1810\,\dfrac{\mathrm m}{\mathrm s}\right)\sin20.0^\circ(9.20\,\mathrm s)+\dfrac12a_y(9.20\,\mathrm s)^2\implies a_y=110\,\dfrac{\mathrm m}{\mathrm s^2}[/tex]
The acceleration vector then has direction [tex]\theta[/tex] where
[tex]\tan\theta=\dfrac{a_y}{a_x}\implies\theta=79.2^\circ[/tex]
