Solve the following :

For last $6 \mathrm{~m}$,

$\mathrm{t}=0.2 \mathrm{sec} ; \mathrm{s}=6 \mathrm{~m} ; \mathrm{a}=\mathrm{g}$

$s=u t+\frac{1}{2} a t^{2}$

$6=\mathrm{u}(0.2)+\frac{\frac{1}{2}}{}(\mathrm{~g})(0.2)^{2}$

$\mathrm{u}=29 \mathrm{~m} / \mathrm{s}$

Before last $6 \mathrm{~m}$,

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

$(29)^{2}=0^{2}+2(g) s$

$\mathrm{S}_{1}=42 \mathrm{~m}$

Total distance $=42+6=48 \mathrm{~m}$