MHT CET · Physics · Motion In Two Dimensions
An aeroplane is flying in a horizontal direction with a velocity of \(540 \mathrm{~km} / \mathrm{hr}\) at a height of \(1960 \mathrm{~m}\). When it is vertically above the point A on the ground, a body is dropped from it. The body strikes the ground at point \(\mathrm{B}\). The distance \(\mathrm{AB}\) is equal to
\(\left(\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^2\right)\)
- A \(2000 \mathrm{~m}\)
- B \(3000 \mathrm{~m}\)
- C \(3600 \mathrm{~m}\)
- D \(4000 \mathrm{~m}\)
Answer & Solution
Correct Answer
(B) \(3000 \mathrm{~m}\)
Step-by-step Solution
Detailed explanation
From \(\mathrm{h}=\frac{1}{2} \mathrm{gt}^2\)
We have \(\mathrm{t}_{\mathrm{OB}}=\sqrt{\frac{2 \mathrm{~h}_{\mathrm{OB}}}{\mathrm{g}}}=\sqrt{\frac{2 \times 1960}{9.8}}=20 \mathrm{~s}\)
Horizontal distance \(\mathrm{AB}=\mathrm{vt}_{\mathrm{OB}}=\left(540 \times \frac{5}{18} \mathrm{~m} / \mathrm{s}\right)(20 \mathrm{~s})=3000 \mathrm{~m}\)
We have \(\mathrm{t}_{\mathrm{OB}}=\sqrt{\frac{2 \mathrm{~h}_{\mathrm{OB}}}{\mathrm{g}}}=\sqrt{\frac{2 \times 1960}{9.8}}=20 \mathrm{~s}\)
Horizontal distance \(\mathrm{AB}=\mathrm{vt}_{\mathrm{OB}}=\left(540 \times \frac{5}{18} \mathrm{~m} / \mathrm{s}\right)(20 \mathrm{~s})=3000 \mathrm{~m}\)
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