A projectile is launched at an angle '$\alpha$' with the horizontal with a velocity $20 \mathrm{~ms}^{-1}$. After 10 s, its inclination with horizontal is '$\beta$'. The value of $\tan \beta$ will be : $\left(\mathrm{g}=10 \mathrm{~ms}^{-2}\right)$

(B) $\tan \alpha-5 \sec \alpha$

(A) $\tan \alpha+5 \sec \alpha$

(C) $2 \tan \alpha-5 \sec \alpha$

(D) $2 \tan \alpha+5 \sec \alpha$

JEE Main · 2022 · 27 Jun 2022 · Shift 1 · Kinematics

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) $tan α + 5 sec α$
(B) $tan α - 5 sec α$ ✓ Correct
(C) $2 tan α - 5 sec α$
(D) $2 tan α + 5 sec α$

Answer: (B)

Solution

The horizontal component of velocity remains constant:

v_{x} = u_{x} = 20 cos α

.

The vertical component after 10 s is:

v_{y} = 20 sin α - g t = 20 sin α - 10 \times 10 = 20 sin α - 100

.

The inclination

β

at time

t = 10

s is given by:

tan β = \frac{v _{y}}{v _{x}} = \frac{20 sin α - 100}{20 cos α} = \frac{20 sin α}{20 cos α} - \frac{100}{20 cos α} = tan α - 5 sec α

Therefore, the answer is (B).

JEE Main · 2022 · 27 Jun 2022 · Shift 1 · Single correct option

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