b
Let \(h\) be height of the tower and \(t\) is the time taken by the body to reach the ground.

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, \(\left.u=0\right)\)

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

\(40=(2 t-2) g\)

\(t=3 s\)

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

\(h=45 \;m\).