A ‘fishbowl’ of height $\frac{4r}{3}$ is formed by removing the top third of a sphere (of radius r). The fishbowl is fixed in sand so that its rim is parallel with the ground. A small marble of mass m rests at the bottom of the fishbowl. Assuming all surfaces are frictionless and ignoring air resistance, the maximum initial velocity that could be given to the marble for it to land back in the fishbowl is

$$\begin{gathered} r^2={\left(\frac{S_x}{2}\right)}^2+{\left(\frac{r}{3}\right)}^2\to S_x=\frac{4\sqrt{2}}{3}r \\ \sin \theta =\frac{S_x}{2}\times \frac{1}{r}=\frac{2\sqrt{2}}{3} and \cos \theta =\frac{r}{3}\times \frac{1}{r}=\frac{1}{3} \end{gathered}$$

$$\begin{gathered} S_x=\frac{4\sqrt{2}}{3}r S_y=0 \\ {u}_x=v_1\cos \theta u_y=v_1\sin \theta \\ {u}_x=v_1\cos \theta u_y=v_1\sin \theta \\ {v}_x=v_1\cos \theta v_y=? \\ {a}_x=0 a_y=-g \\ t=? t=? \end{gathered}$$

Considering the y direction :
 Substituting in the relevant values gives
$0=\left(v_1\sin \theta \right)t-\frac{g}{2}{t}^2$

And so  $t=\frac{2v_1\sin \theta }{g}$
 Now consider the $x$  direction  $S_x=u_{x}t+\frac{1}{2}{a}_x{t}^2$
 Since $a_x=0$, substituting in values gives
 $\frac{4\sqrt{2}}{3}r=v_1\cos \theta \times \frac{2v_1\sin \theta }{g}$          
  Substituting in values of  $\sin \theta$ and $\cos \theta$  and simplifying eventually yields
   $v_1^2=3gr$
 We are told that the height of the fishbowl is $\frac{4r}{3}$. This leads to the energy conservation equation 
 becoming  $\frac{1}{2}m{v}_0^2+0=\frac{1}{2}m{v}_1^2+mg\times \frac{4r}{3}$
 Substituting in  $v_1^2=3gr$ and simplifying finally yields
$v_0=\sqrt{\frac{17gr}{3}}$