Answer
a) $2.91\;\rm m/s$
b) $14.3\;\rm N$
Work Step by Step
a) First of all, we need to draw a pictorial sketch to find the missing parts (angles and segments) in the given figure, as shown in the figures below.
We found all the angles needed to find $r$ which is the radius of the circular path of the ball so we can find the tension.
To find the speed, we need to apply Newton's second law on the ball.
$$\sum F_y=T\cos30^\circ+T\cos60^\circ-mg=ma_y=m(0)=0$$
Thus,
$$T\cos30^\circ+T\cos60^\circ=mg$$
$$T\left[\cos30^\circ+ \cos60^\circ\right]=mg$$
$$T=\dfrac{mg}{\left[\cos30^\circ+ \cos60^\circ\right]}\tag 1$$
$$\sum F_r=T\sin 30^\circ+T\sin60^\circ =ma_r=m\dfrac{v^2}{r}$$
Thus,
$$T\left[\sin 30^\circ+ \sin60^\circ \right]=m\dfrac{v^2}{r}$$
Plugging $T$ from (1);
$$\dfrac{ \color{red}{\bf\not}mg}{\left[\cos30^\circ+ \cos60^\circ\right]}\left[\sin 30^\circ+ \sin60^\circ \right]= \color{red}{\bf\not}m\dfrac{v^2}{r}$$
$$\dfrac{ g\left[\sin 30^\circ+ \sin60^\circ \right]}{\left[\cos30^\circ+ \cos60^\circ\right]}= \dfrac{v^2}{r}$$
Noting that $\left[\cos30^\circ+ \cos60^\circ\right]=\left[\sin 30^\circ+ \sin60^\circ \right]$
$$v=\sqrt{\dfrac{ gr\left[\sin 30^\circ+ \sin60^\circ \right]}{\left[\cos30^\circ+ \cos60^\circ\right]} } =\sqrt{gr} $$
$$v=\sqrt{gr} \tag 2$$
Now we need to find which $r$ is given by
$$\tan 30^\circ=\dfrac{r}{1+h}$$
Thus,
$$r=(1+h)\tan 30^\circ\tag 3$$
We can see from the lower right triangle that $h$ is given by $$\tan30^\circ=\dfrac{h}{r}$$
And hence, $$h=r\tan30^\circ$$
Plugging into (3);
$$r=(1+r\tan30^\circ)\tan 30^\circ=\tan 30^\circ+r\tan^230^\circ $$
Thus,
$$r-r\tan^230^\circ=\tan 30^\circ$$
$$r=\dfrac{\tan 30^\circ}{1-\tan^230^\circ}$$
Plugging into (2);
$$v=\sqrt{g\dfrac{\tan 30^\circ}{1-\tan^230^\circ}} $$
$$v=\sqrt{9.8\times \dfrac{\tan 30^\circ}{1-\tan^230^\circ}} $$
$$v=\color{red}{\bf 2.91}\;\rm m/s $$
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b) The tension is given by (1);
$$T=\dfrac{mg}{\left[\cos30^\circ+ \cos60^\circ\right]}=\dfrac{2\times 9.8}{\left[\cos30^\circ+ \cos60^\circ\right]}$$
$$T=\color{red}{\bf 14.3}\;\rm N $$
