Answer
a) $3.85\;\rm m/s$
b) $19.1\;\rm m/s^2,\;14.8^\circ$
Work Step by Step
First, we need to draw the force diagram of the ball when its wire is making an angle of 30$^\circ$ with the vertical, as we see in the figures below.
Now we need to apply Newton's second law where $x$-direction is the radial direction and $z$-direction is the tangential direction.
$$\sum F_x=T+mg\cos\theta=ma_r=m\dfrac{v^2}{R}$$
Thus,
$$T+mg\cos\theta =m\dfrac{v^2}{R}$$
Solving for $v$;
$$\dfrac{T+mg\cos\theta }{m}R= v^2 $$
$$v=\sqrt{\dfrac{T+mg\cos\theta }{m}R}$$
Plugging the known;
$$v=\sqrt{\dfrac{20+(2\times9.8\cos30^\circ)}{2}\times 0.8}$$
Thus,
$$v=\color{red}{\bf 3.85}\;\rm m/s$$
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b) To find the acceleration of the ball at the same place, we need to find its tangential acceleration and its radial acceleration.
$$a_r=\dfrac{v^2}{R}$$
Plugging from above;
$$a_r=\dfrac{3.85^2}{0.8}=\bf 18.5\;\rm m/s^2 $$
$$\sum F_z=\color{red}{\bf\not} mg\sin\theta=\color{red}{\bf\not} ma_T$$
Thus,
$$a_T=g\sin\theta=9.8\times \sin30^\circ=\bf 4.9\;\rm m/s^2$$
Hence, the acceleration of the ball is given by applying the Pythagorean theorem.
$$a=\sqrt{a_T^2+a_r^2}=\sqrt{4.9^2+18.5^2}$$
$$a=\color{red}{\bf19.14}\;\rm m/s^2$$
And its direction relative to the rope (the $x$-direction) is given by
$$\tan\theta_1= \dfrac{a_T}{a_r}$$
$$\theta_1=\tan^{-1}\left[ \dfrac{a_T}{a_r}\right]$=\tan^{-1}\left[ \dfrac{4.9}{18.5}\right]$$
$$\theta_1=\color{red}{\bf14.8^\circ}$$
