Answer
$a).$ (3) four times that of cliff B
$b).$ Height of cliff A = 15.88m, and that of cliff B = 3.97m.
Work Step by Step
a). $t_{a}=2\times t_{b}$
$S=ut+\frac{1}{2}gt^{2}=0+\frac{1}{2}gt^{2}=\frac{1}{2}gt^{2}$
So, $S_{a}=\frac{1}{2}gt_{a}^{2} \,\,and \,\, S_{b}=\frac{1}{2}gt_{b}^{2}$
So, $\frac{S_{a}}{S_{b}}=(\frac{t_{a}}{t_{b}})^{2}=4$
$S_{a}=4\times S_{b}$
So, the answer is (3)
b). $S_{a}=0.5\times 9.8\times 1.8^{2}=15.88m$
and $S_{b}=3.97m$
