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JEE Mains · Physics · STD 11 - 3.2 motion in plane

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 : \(\left( g =10 \; ms ^{-2}\right)\)

  1. A \(\tan \alpha+5 \sec \alpha\)
  2. B \(\tan \alpha-5 \sec \alpha\)
  3. C \(2 \tan \alpha-5 \sec \alpha\)
  4. D \(2 \tan \alpha+5 \sec \alpha\)
Verified Solution

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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