Answer
$\theta=43.3^o$
Work Step by Step
The pendulum is given initial velocity $v_0=2m/s$ and moves until it comes to a halt at $v_f=0$. During this period, it goes upward, meaning its KE decreases while PE increases. From the principle of energy conservation, we have $$\Delta KE = -\Delta PE$$ $$\frac{1}{2}m\Delta v^2=-mg\Delta h$$ $$\frac{1}{2} (v_f^2-v_0^2)=-g\Delta h$$
From here, we can find how much the pendulum has gone upward. We have $$\Delta h=\frac{v_f^2-v_0^2}{-2g}=0.204m$$
From the figure below, we see that we can find $\theta$ using its cosine. $$\cos\theta=\frac{0.75-\Delta h}{0.75}=0.728$$ $$\theta=43.3^o$$
