Answer
$\dfrac{3\pi}{4}$
Work Step by Step
Since we know, velocity $v(t)=r'(t)=\lt \dfrac{\sqrt 2}{2}, \dfrac{\sqrt 2}{2}-32t,0 \gt$
and $v(0)=\lt \dfrac{\sqrt 2}{2}, \dfrac{\sqrt 2}{2},0 \gt=\dfrac{1}{\sqrt 2}\lt 1,1,0 \gt$
As we know acceleration $a(t)=v'(t)=\lt 0,-32,0 \gt$
and $a(0)=32 \lt 0,-1,0 \gt$
Hence, $\theta =\cos ^{-1}\dfrac{v(0) \cdot a(0)}{|v(0)||a(0)|}=\cos ^{-1}(\dfrac{0-1+0}{\sqrt{1^2+1^2+0^2}\sqrt{0^2+(-1)^2+0^2}})=\cos ^{-1} \dfrac{1}{\sqrt 2}$
or, $\theta=\dfrac{3\pi}{4}$
