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 : \(\left( g =10 \; ms ^{-2}\right)\)
- A \(\tan \alpha+5 \sec \alpha\)
- B \(\tan \alpha-5 \sec \alpha\)
- C \(2 \tan \alpha-5 \sec \alpha\)
- D \(2 \tan \alpha+5 \sec \alpha\)
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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