So, it is accelerating $\therefore a=\frac{v^{2}}{R}$

For vertical direction $a=0$

then, according to question to newton's law $F=m g=m a$

$f_{i}=\frac{m v^{2}}{R} \quad...(i)$

In case of no skidding. $f_{s}=\mu_{s} F=\mu_{s} m g...(i i)$

$\therefore$ from $(i) \&(i i)$ we get

$\mu_{s} m g=\frac{m r^{2}}{R}$

$o r, v^{2}=\mu_{s} g R$

$v=\sqrt{2.0 \times 10 \times 45}$

$=30 \mathrm{m} / \mathrm{s}$