Answer
$\cos ^{-1} (\dfrac{1}{\sqrt 2})$ or, $\dfrac{3\pi}{4}$
Work Step by Step
We know that the velocity and acceleration is defined as: velocity is given as: $v(t)=r'(t)$ and acceleration is given as: $a(t)=v'(t)$
Now, $v(t)=\lt \dfrac{\sqrt 2}{2}, \dfrac{\sqrt 2}{2}-32(t),0 \gt \\ \implies v(0)=\lt \dfrac{\sqrt 2}{2}, \dfrac{\sqrt 2}{2},0 \gt=\dfrac{1}{\sqrt 2}\lt 1,1,0 \gt$
and $a(t)=v'(t)=\lt 0,-32,0 \gt \\ \implies a(0)=32 \lt 0,-1,0 \gt$
As we know that $\theta =\cos ^{-1}\dfrac{v(0) \cdot a(0)}{|v(0)||a(0)|}$
Thus, $\theta =\cos ^{-1}(\dfrac{(0-1+0)}{\sqrt{1^2+1^2+0^2}\sqrt{0^2+(-1)^2+0^2}})=\cos ^{-1} (\dfrac{1}{\sqrt 2})=\dfrac{3\pi}{4}$
