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JEE Main 2007
Exam
JEE Main
Year
2007
Subject
Physics
Chapter
Motion in Two Dimensions

JEE Main 2007 · Physics · Motion in Two Dimensions

A particle just clears a wall of height b at distance a and strikes...

A particle just clears a wall of height bb at distance aa and strikes the ground at a distance cc from the point of projection. The angle of projection isA particle just clears a wall of height b at distance a and strikes the ground at a distance c from the point of

Options

  1. Atan⁡−1bac\tan ^{-1} \frac{b}{a c}
  2. B45∘45^{\circ}
  3. Ctan⁡−1bca(c−a)\tan ^{-1} \frac{b c}{a(c-a)}
  4. Dtan⁡−1bca\tan ^{-1} \frac{b c}{a}

Correct answer

C. tan⁡−1bca(c−a)\tan ^{-1} \frac{b c}{a(c-a)}

Step-by-step solution

a=(ucos⁡α)ta=(u \cos \alpha) t and b=(usin⁡α)t−12gt2b=(u \sin \alpha) t-\frac{1}{2} g t^2 b=atan⁡α−12ga2u2cos⁡2αb=a \tan \alpha-\frac{1}{2} g \frac{a^2}{u^2 \cos ^2 \alpha} also, c=u2sin⁡2αgc=\frac{u^2 \sin 2 \alpha}{g} b=atan⁡α−a2g2(sin⁡2αcg)sec⁡2αb=a \tan \alpha-\frac{a^2 g}{2}\left(\frac{\sin 2 \alpha}{c g}\right) \sec ^2 \alpha b=atan⁡α−a22c2tan⁡αb=a \tan \alpha-\frac{a^2}{2 c} 2 \tan \alpha ⇒(a−a2c)tan⁡α=b\Rightarrow\left(a-\frac{a^2}{c}\right) \tan \alpha=b tan⁡α=bca(c−a)\tan \alpha=\frac{b c}{a(c-a)}. A particle just clears a wall of height b at distance a and strikes the ground at a distance c from the point of
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