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JEE Main 202225 JuneEvening Shift
Exam
JEE Main
Year
2022
Date
25 June
Shift
Evening Shift
Subject
Physics
Chapter
Motion in Two Dimensions

JEE Main 2022 · 25 June · Evening Shift · Physics · Motion in Two Dimensions

Given below are two statements. One is labelled as Assertion A and...

Given below are two statements. One is labelled as Assertion A and the other is labelled as Reason R.

Assertion A: Two identical balls AA and BB thrown with same velocity 'uu' at two different angles with horizontal attained the same range RR. If A and BB reached the maximum height h1{h}_{1} and h2{h}_{2} respectively, then R=4h1h2R = 4\sqrt{{h}_{1}{h}_{2}}

Reason R: Product of said heights. h1h2=(u2sin⁡2θ2g)⋅(u2cos⁡2θ2g){h}_{1}{h}_{2} = \left(\dfrac{{u}^{2}\sin^{2}\theta}{2g}\right) \cdot \left(\dfrac{{u}^{2}\cos^{2}\theta}{2g}\right)

Options

  1. ABoth A and R are true and R is the correct explanation of A.
  2. BBoth A and R are true but R is NOT the correct explanation of A.
  3. CA is true but R is false.
  4. DA is false but R is true.

Correct answer

A. Both A and R are true and R is the correct explanation of A.

Step-by-step solution

Range of two balls are R1=u2sin⁡2θ1g=R2=u2sin⁡2θ2g{R}_{1} = \dfrac{{u}^{2}\sin 2{\theta}_{1}}{g} = {R}_{2} = \dfrac{{u}^{2}\sin 2{\theta}_{2}}{g}.

Now, maximum height by both balls are h1=u2sin⁡2θ12g{h}_{1} = \dfrac{{u}^{2}\sin^{2}{\theta}_{1}}{2g} and h2=u2cos⁡2θ22g{h}_{2} = \dfrac{{u}^{2}\cos^{2}{\theta}_{2}}{2g}

∴h1h2=(u2sin⁡2θ2g)(u2cos⁡2θ2g)=u2sinθcosθ2g∴ \sqrt{{h}_{1}{h}_{2}} = \left(\dfrac{{u}^{2}\sin^{2}\theta}{2g}\right)\left(\dfrac{{u}^{2}\cos^{2}\theta}{2g}\right) = \dfrac{{u}^{2}sin\theta cos\theta}{2g}=R4= \dfrac{R}{4}

⇒R=4h1h2{⇒ R = 4\sqrt{{h}_{1}{h}_{2}}}^{}

Hence, both A and R are true and R is the correct explanation of A.

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