JEE Mains · Physics · STD 11 - 2. motion in straight line
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
- A \(6.5\)
- B \(8\)
- C \(10.3\)
- D \(5.4\)
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 \(\mathrm{t}+\frac{1}{2} \mathrm{sec}\) \(81=\frac{1}{2} \times \mathrm{a} \times \mathrm{t}^{2} \quad \Rightarrow \mathrm{t}=9 \sqrt{\frac{2}{\mathrm{a}}}\)…
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