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Physics · Motion in one Dimension

JEE Main 2024 — 6 April, Shift 2 — Question 50

A body projected vertically upwards with a certain speed from the top of a tower reaches the ground in t1t_{1}. If it is projected vertically downwards from the same point with the same speed, it reaches the ground in t2t_{2}. Time required to reach the ground, if it is dropped from the top of the tower, is :

  1. Option A:

    t1t2\sqrt{\mathrm{t}_{1} \mathrm{t}_{2}}

    Correct
  2. Option B:

    t1−t2\sqrt{t_{1}-t_{2}}

  3. Option C:

    t1t2\sqrt{\frac{t_{1}}{t_{2}}}

  4. Option D:

    t1+t2\sqrt{t_{1}+t_{2}}

Answer: A

Step-by-step solution

t1=u+u2+2ghg\mathrm{t}_{1}=\frac{\mathrm{u}+\sqrt{\mathrm{u}^{2}+2 \mathrm{gh}}}{\mathrm{g}}

t2=−u+u2+2ghg\mathrm{t}_{2}=\frac{-\mathrm{u}+\sqrt{\mathrm{u}^{2}+2 \mathrm{gh}}}{\mathrm{g}}

t=2ghg\mathrm{t}=\frac{\sqrt{2 \mathrm{gh}}}{\mathrm{g}} t1t2=(u2+2gh)−u2 g2=2ghg2=t2\mathrm{t}_{1} \mathrm{t}_{2}=\frac{\left(\mathrm{u}^{2}+2 \mathrm{gh}\right)-\mathrm{u}^{2}}{\mathrm{~g}^{2}}=\frac{2 \mathrm{gh}}{\mathrm{g}^{2}}=\mathrm{t}^{2}

⇒t=t1t2\Rightarrow \mathrm{t}=\sqrt{\mathrm{t}_{1} \mathrm{t}_{2}}

Answer key and solution verified before publishing.

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

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