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A ball is dropped from the top of a $100\; m$ high tower on a planet. In the last $\frac{1}{2}\;s $ before hitting the ground, it covers a distance of $19\; \mathrm{m}$. Acceleration due to gravity (in $\mathrm{ms}^{-2}$ ) near the surface on that planet is
$6.5$
$8$
$10.3$
$5.4$
Solution
Time to travel $81 \mathrm{m}$ is t sec.
Time to travel $100 \mathrm{m}$ is $\mathrm{t}+\frac{1}{2} \mathrm{sec}$
$81=\frac{1}{2} \times \mathrm{a} \times \mathrm{t}^{2} \quad \Rightarrow \mathrm{t}=9 \sqrt{\frac{2}{\mathrm{a}}}$
$100=\frac{1}{2} \times \mathrm{a} \times\left(\mathrm{t}+\frac{1}{2}\right)^{2} \Rightarrow \mathrm{t}+\frac{1}{2}=10 \sqrt{\frac{2}{\mathrm{a}}}$
$9 \sqrt{\frac{2}{a}}+\frac{1}{2}=10 \sqrt{\frac{2}{a}}$
$\frac{1}{2}=\sqrt{\frac{2}{a}}$
$a=8 \mathrm{m} / \mathrm{s}^{2}$
