JEE Main · 2022 · 24 Jun 2022 · Shift 1 · Kinematics

A projectile is projected with velocity of

25 m / s

at an angle

θ

with the horizontal. After t seconds its inclination with horizontal becomes zero. If R represents horizontal range of the projectile, the value of

θ

will be : [use

g = 10 m / s^{2}

]

Answer: (D)

Solution

At the highest point (inclination becomes zero), time

t = \frac{V s i n θ}{g}

, so

V = \frac{g t}{s i n θ}

.

The range

R = \frac{V ^{2} s i n 2 θ}{g} = \frac{g ^{2} t ^{2}}{s i n ^{2} θ} \cdot \frac{2 s i n θ c o s θ}{g} = \frac{2 g t ^{2} c o s θ}{s i n θ}

Therefore

tan θ = \frac{2 g t ^{2}}{R} = \frac{20 t ^{2}}{R}

Hence

θ = cot^{- 1} (\frac{R}{20 t ^{2}})

JEE Main · 2022 · 24 Jun 2022 · Shift 1 · Single correct option