If a ball is thrown vertically with speed $u$, the distance covered during the last $t$ seconds of its ascent is:

• A

  $u t$
• Correct answer
  B

  $\frac{1}{2} g t^2$
• C

  $u t-\frac{1}{2} g t^2$
• D

  $(u+g t) t$

Solution

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

Hence, the correct answer is option: 'B'.