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

If a ball is thrown vertically with speed $u$, the distance covered during the last $t$ seconds of its ascent is:
  1. $u t$
  2. $\frac{1}{2} g t^2$
  3. $u t-\frac{1}{2} g t^2$
  4. $(u+g t) t$

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 $\mathrm{B}=V_B=u-g$ $(T-t)$
$=u-g\left(\frac{u}{g}-t\right)=g t$

$\Rightarrow h=V_B t-\frac{1}{2} g t^2$

Asked in: NEET 2003

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