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JEE Mains · Physics · STD 11 - Motion in plane
  • JEE Mains 2022
  • Hard
  • MCQ

A projectile is launched at an angle ' \(\alpha\) ' with the horizontal with a velocity \(20 \; ms ^{-1}\). After \(10 s\), its inclination with horizontal is ' \(\beta\) '. The value of \(\tan \beta\) will be : \(\left( g =10 \; ms ^{-2}\right)\)

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Answer & Solution

Correct Answer

(B) \(\tan \alpha-5 \sec \alpha\)

Step-by-step Solution

Detailed explanation

\(v_{x}=u_{x}=20 \cos \alpha\)

\(v_{y}=20 \sin \alpha-10 \times 10\)

\(\tan \beta=\frac{v_{y}}{v_{x}}=\frac{20 \sin \alpha-100}{20 \cos \alpha}\)

\(=\tan \alpha-5 \sec \alpha\)

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