Initially body is at rest.

$\therefore$ $h = {1 \over 2}g{t^2}$

$\Rightarrow$ $g = {{2h} \over {{t^2}}}$

Maximum percentage error,

${{\Delta g} \over g} \times 100 = \left[ {{{\Delta h} \over h} + 2{{\Delta t} \over t}} \right] \times 100$

Given that
${{{\Delta h} \over h} \times 100}$ = e₁

and ${{{\Delta t} \over t} \times 100}$ = e₂

$\therefore$ ${{\Delta g} \over g} \times 100$ = e₁ + 2e₂