Here, $u =0, a = g$

$h=u t+\frac{1}{2} g t^{2}$ or $h=0 \times t+\frac{1}{2} g t^{2}$

$h=\frac{1}{2} g t^{2}$

Distance covered in last two seconds is

$40=\frac{1}{2} g t^{2}-\frac{1}{2} g(t-2)^{2}$ (Here, $\left.u=0\right)$

$40=\frac{1}{2} g t^{2}-\frac{1}{2} g\left(t^{2}+4-4 t\right)$

$40=(2 t-2) g$

$t=3 s$

$h=\frac{1}{2} \times 10 \times(3)^{2}$

$h=45 \;m$.