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If a ball is thrown vertically with speed $u$, the distance covered during the last $t$ seconds of its ascent is:
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Verified Answer
The correct answer is:
$\frac{1}{2} g t^2$
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 $\mathrm{B}=V_B=u-g$ $(T-t)$
$=u-g\left(\frac{u}{g}-t\right)=g t$
$\Rightarrow h=V_B t-\frac{1}{2} g t^2$
Again Let $h$ be the distance covered during last ' $t$ ' second of its ascent.
$\therefore$ Velocity at point $\mathrm{B}=V_B=u-g$ $(T-t)$
$=u-g\left(\frac{u}{g}-t\right)=g t$
$\Rightarrow h=V_B t-\frac{1}{2} g t^2$
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