JEE Mains · Physics · STD 11 - Motion in straight line
- JEE Mains 2020
- Medium
- Numerical
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
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(8\)
Step-by-step Solution
Detailed explanation
Time to travel \(81 \mathrm{m}\) is t sec.
Time to travel \(100 \mathrm{m}\) is \(\mathrm{t}+\frac{1}{2} \mathrm{sec}\)
\(81=\frac{1}{2} \times \mathrm{a} \times \mathrm{t}^{2} \quad \Rightarrow \mathrm{t}=9 \sqrt{\frac{2}{\mathrm{a}}}\)
\(100=\frac{1}{2} \times \mathrm{a} \times\left(\mathrm{t}+\frac{1}{2}\right)^{2} \Rightarrow \mathrm{t}+\frac{1}{2}=10 \sqrt{\frac{2}{\mathrm{a}}}\)
\(9 \sqrt{\frac{2}{a}}+\frac{1}{2}=10 \sqrt{\frac{2}{a}}\)
\(\frac{1}{2}=\sqrt{\frac{2}{a}}\)
\(a=8 \mathrm{m} / \mathrm{s}^{2}\)
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