A projectile is launched at an angle ' α ' with horizontal with a velocity 20 ms ^{-1} . Af

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3-2.Motion in Plane

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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

$2 \tan \alpha+5 \sec \alpha$

(JEE MAIN-2022)

Solution

$v_{x}=u_{x}=20 \cos \alpha$

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

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

$=\tan \alpha-5 \sec \alpha$

Standard 11

Physics

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