• # question_answer If a ball is thrown vertically upwards with speed u, the distance covered during the last T seconds of its ascent is :- A)  ut                               B)  $\frac{1}{2}g{{t}^{2}}$C)  $ut-\frac{1}{2}g{{t}^{2}}$                         D)  $(u+gt)t$

Let time of flight be T then $T=\frac{u}{g}$ Let h be the distance covered during last 't' second of its ascent               Velocity at point $B={{v}_{B}}=u-g(T-t)$             $=u-g\left( \frac{u}{g}-t \right)=gt$ $\Rightarrow$               $h={{v}_{B}}t-\frac{1}{2}g{{t}^{2}}$ $\Rightarrow$               $h=g{{t}^{2}}-\frac{1}{2}g{{t}^{2}}=\frac{1}{2}g{{t}^{2}}$