A body is thrown up with a velocity  $100m{s}^{-1}$. It travels 5m in the last second of its journey. If the same body is thrown up with a velocity  $200m{s}^{-1}$ , how much distance (in meter) will it travel in the last second  $(g=10m{s}^{-2})$

$S_{n^{th}}=4+\frac{a}{2}(2n-1)$

$u=100m{s}^{-1},a=-10m{s}^{-2}$  and  $s=5 m$

$5=100-5(2n-1)$  gives  $n=10\text{ s}$  

Body when thrown up with velocity   $200{\text{ ms}}^{-1}$ will take 20s to reach the highest point. Distance travelled in 20^th second is  $200-5(200\times 2-1)=5m$  in the last second of upward  journey, the bodies will travel same distance.