JEE Main 2022 · 27 Jun · Shift 1 · Physics Q31

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PhysicsSingle correct+4 / -1Question 31 of 90

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 : ( $g=10$ ms $^{-2}$ ).

\tan\alpha+5\sec\alpha

\tan\alpha-5\sec\alpha

2\tan\alpha-5\sec\alpha

2\tan\alpha+5\sec\alpha

Answer and worked solution

Official answer: B

Velocity after 10 s. $v_x=20\cos\alpha$ , $v_y=20\sin\alpha-100$ .

Slope. $\tan\beta=\dfrac{20\sin\alpha-100}{20\cos\alpha}=\tan\alpha-5\sec\alpha$ .

(B)

Topics:Projectile motion

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JEE Main 2022 · 27 Jun · Shift 1, Physics question 31