A person travelling at

$\mathrm{u}=43.2^{\times \frac{5}{18}}=12 \mathrm{~m} / \mathrm{s} ; \mathrm{v}=0 ; \mathrm{a}=-6 \mathrm{~m} / \mathrm{s}^{2}$

Using,

$v^{2}=u^{2}+2 a s$

$\mathrm{O}^{2}=(12)^{2}-2^{\mathrm{X}_{6}} \mathrm{X}_{\mathrm{s}}$

$s=12 m$