Answer
$a).$ Time taken on moon will be $(1) \sqrt 6$ times it would take on earth.
$b).$ On earth, maximum height$=16.53m$, total time$=3.68s$.
On moon, maximum height$=99.18m$, total time$=22.04s$.
Work Step by Step
Let, Acceleration due to gravity on earth = $g_{e}$
Acceleration due to gravity on moon = $g_{m}$
Now, $g_{m}=\frac{1}{6}g_{e}$
If height=S,
$S=ut+\frac{1}{2}gt^{2}$
$S=0+\frac{1}{2}g_{e}t_{e}^{2}$ - (1)
$S=0+\frac{1}{2}g_{m}t_{m}^{2}$ - (2)
So, equating the two above equations,
$\frac{t_{m}}{t_{e}}=\sqrt \frac{g_{e}}{g_{m}}=\sqrt 6$
a). Time taken on moon will be $(1) \sqrt 6$ times it would take on earth.
b).
$v^{2}=u^{2}+2aS$
$v=u+at$
On earth, $0=18-9.8t$
$t=1.84s$ Total time$=2\times 1.84=3.68s$
Also, $0=18^{2}-2\times 9.8\times S$
$S=16.53m$
On moon, $0=18-\frac{9.8}{6}t$
$t=11.02s$ Total time$=22.04s$
Also, $0=18^{2}-2\times \frac{9.8}{6}\times S$
$S=99.18m$
