Answer
$tan^{-1}{\frac{2sin\theta}{(cos\theta-1)}}$
Work Step by Step
Please see the attached image first.
Let's assume the angle between the acceleration $(\vec{a})$ vector and the original direction of the object's motion is '$\alpha$'
Now we apply equation $V=u+at$ to the object
$\rightarrow V=u+at$
Let's plug known values into this equation.
$2Vcos\theta=V+acos\alpha\space t$
$\frac{V}{t}(2cos\theta-1)= acos\alpha-(1)$
$\downarrow V=u+at$
Let's plug known values into this equation.
$2Vsin\theta=0+asin\alpha\space t$
$\frac{2Vsin\theta}{t}= asin\alpha-(2)$
$(2)\div(1)=\gt$
$tan\alpha= \frac{\frac{2Vsin\theta}{t}}{\frac{V}{t}(2cos\theta-1)}=\gt \alpha = tan^{-1}\frac{2sin\theta}{(cos\theta-1)}$
