A projectile is launched at an angle ' α ' with the horizontal with a velocity 20 ms ⁻¹. After 10 s, its inclination with horizontal is ' β '. The value of tan β will be: (g = 10 ms ⁻²). (a) tan α + 5sec α (b) tan α - 5sec α (c) 2tan α - 5sec α (d) 2tan α + 5sec α

Answer: (b). vx = 20cos α, vy = 20sin α - 10 × 10 = 20sin α - 100 tan β = (vy)/(vx) = (20sin α - 100)/(20cos α) tan β = tan α - 5sec α So the answer is (b).

A projectile is launched at an angle ' α ' with the horizontal with a… | JEE Main 2022 (27 Jun, Shift 1)

Q12JEE Main 2022 · 27 JUN · SHIFT 1Medium

Velocity, Momentum & Energy in Flight

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 ${}^{-2}$ ).

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Answer: (b)

$v_x = 20\cos\alpha$ , $v_y = 20\sin\alpha - 10 \times 10 = 20\sin\alpha - 100$
$\tan\beta = \frac{v_y}{v_x} = \frac{20\sin\alpha - 100}{20\cos\alpha}$
$\tan\beta = \tan\alpha - 5\sec\alpha$
So the answer is (b).

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