Correct option (b) \(45 \mathrm{~m}\)
Explanation:
Let \(h\) be height of the tower and \(t\) is the time taken by the body to reach the ground.
Here, \(u=0, a=g\)
\(\therefore \quad h=u t+\frac{1}{2} g t^{2}\)
or \(h=0 \times t+\frac{1}{2} g t^{2}\)
or \(h=\frac{1}{2} g t^{2}\) (i)
Distance covered in last two seconds is
\(40=\frac{1}{2} g t^{2}-\frac{1}{2} g(t-2)^{2} \quad(\) Here, \(u=0)\)
or \(40=\frac{1}{2} g t^{2}-\frac{1}{2} g\left(t^{2}+4-4 t\right)\)
or \(40=(2 t-2) g\) or \(t=3 \mathrm{~s}\)
From eqn (i), we get
\(h=\frac{1}{2} \times 10 \times(3)^{2}\) or \(h=45 \mathrm{~m}\)
