A ball is dropped from height h=100 m, from surface of planet . If in the last  $\frac{1}{2}$s of its journey, it covers a distance of 19 m. Then acceleration due to gravity (in $m{s}^{-2}$)  on that planet is -----------

$$\begin{gathered} Time taken to fall on ground=t=\sqrt{\frac{2h}{a}}=\sqrt{\frac{2\left(100\right)}{a}}=\sqrt{\frac{200}{a}}\mathrm{.....}\left(1\right) \\ Time taken to cover initial dis\tan ce of (100-19 =81m) \\ t\text{'}=\sqrt{\frac{2h\text{'}}{a}} \\ \Rightarrow t-\frac{1}{2}=\sqrt{\frac{2\left(81\right)}{a}}\mathrm{.....}\left(2\right) \end{gathered}$$

  $$\begin{gathered} dividing \left(1\right) by \left(2\right) \\ \frac{t}{t-\frac{1}{2}}=\sqrt{\frac{100}{81}} \\ \Rightarrow \frac{t}{t-\frac{1}{2}}=\frac{10}{9} \\ \Rightarrow 9t=10t-5 \\ \Rightarrow t=5 \\ Substituting in \left(1\right) and then squaring it \\ \Rightarrow 25=\frac{200}{a} \\ \Rightarrow a=8 m/s^2 \end{gathered}$$