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Physics · Motion in one Dimension

JEE Main 2024 — 27 January, Shift 2 — Question 53

A body falling under gravity covers two points A and B separated by 80 m in 2 s . The distance of upper point A from the starting point is m(usex m (use x{\rm{g}} = 10{\rm{;m}}{{\rm{s}}^{ - 2}}$ )

Answer: 45

Numerical answer — enter this value.

Step-by-step solution

From A→B\mathrm{A} \rightarrow \mathrm{B}

−80=−v1t−12×10t2-80=-\mathrm{v}_{1} \mathrm{t}-\frac{1}{2} \times 10 \mathrm{t}^{2}

−80=−2v1−12×10×22-80=-2 \mathrm{v}_{1}-\frac{1}{2} \times 10 \times 2^{2}

−80=−2v1−20-80=-2 \mathrm{v}_{1}-20

−60=−2v1-60=-2 \mathrm{v}_{1}

v1=30 m/s\mathrm{v}_{1}=30 \mathrm{~m} / \mathrm{s}

From O to A v2=u2+2gSv^{2}=u^{2}+2 g S

302=0+2×(−10)(−S)30^{2}=0+2 \times(-10)(-S)

900=20 S900=20 \mathrm{~S}

S=45 m\mathrm{S}=45 \mathrm{~m}

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
27 January, Shift 2
Subject
Physics
Chapter
Motion in one Dimension
Topic
Motion Under Gravity

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