A uniformly charged ring of radius $3a$ and total charge $q$ is placed in $xy-$ plane centered at origin. A point charge $q$ is moving towards the ring along the $z-$ axis and has speed $v$ at $z = 4a$. The minimum value of $v$ such that it crosses the origin is
JEE MAIN 2019, Diffcult
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$\mathrm{U}_{\mathrm{i}}+\mathrm{K}_{\mathrm{i}}=\mathrm{U}_{\mathrm{f}}+\mathrm{K}_{\mathrm{f}}$
$\frac{\mathrm{kq}^{2}}{\sqrt{16 \mathrm{a}^{2}+9 \mathrm{a}^{2}}}+\frac{1}{2} \mathrm{mv}^{2}=\frac{\mathrm{kq}^{2}}{3 \mathrm{a}}$
$\frac{1}{2} m v^{2}=\frac{k q^{2}}{a}\left(\frac{1}{3}-\frac{1}{5}\right)=\frac{2 k q^{2}}{15 a}$
$v=\sqrt{\frac{4 k q^{2}}{15 m a}}$
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