Answer
$22.2\;\rm rpm$
Work Step by Step
We assume that the merry-go-round platform is rotating without friction which means that the angular momentum is constant. We chose the system to be the merry-go-round plus John.
John, before he jumps on the rotating platform merry-go-round, has a constant angular velocity tangential to the merry-go-round.
$$L_i=L_f$$
$$L_{iM}+L_{iJ} =L_f\tag 1$$
$_M\rightarrow$ merry-go-round platform.
$_J\rightarrow$ John.
The initial angular momentum of John is given by $$L_{iJ}=m_Jv_{iJ}R\sin90^\circ=m_Jv_{iJ}R\tag 2$$
($90^\circ$ since he is moving tangent to the platform which makes his velocity angle to the radius is 90 degrees.).
And the initial angular momentum of the platform is given by $$L_{iM}=I_{M}\omega_{iM}=\frac{1}{2}m_MR^2\omega_{iM}\tag 3$$
The final angular momentum is given by
$$L_f=I_{sys}\omega_{f,sys}=\left[\frac{1}{2}m_MR^2+m_JR^2\right]\omega_f\tag 4$$
$_{sys}\rightarrow $ System.
Plugging (2), (3), and (4) into (1):
$$\frac{1}{2}m_MR^2\omega_{iM}+m_Jv_{iJ}R=\left[\frac{1}{2}m_MR^2+m_JR^2\right]\omega_f$$
Hence,
$$\omega_f=\dfrac{\frac{1}{2}m_MR^{\color{red}{\bf\not}2}\omega_{iM}+m_Jv_{iJ}\color{red}{\bf\not}R}{\frac{1}{2}m_MR^{\color{red}{\bf\not}2}+m_JR^{\color{red}{\bf\not}2}}$$
$$\omega_f=\dfrac{\frac{1}{2}m_MR \omega_{iM}+m_Jv_{iJ} }{\frac{1}{2}m_MR +m_JR }$$
Pluggign the known;
$$\omega_f=\dfrac{\frac{1}{2}(250)(1.5)(20)(\frac{2\pi}{60})+(30)(5) }{ \frac{1}{2}(250)(1.5)+(30)(1.5) }=\bf 2.33\;\rm rad/s$$
Therefroe,
$$\omega_f=2.33\;\rm rad/s \left(\frac{60 }{2\pi} \right)=\color{red}{\bf 22.2}\;\rm rpm$$
