JEE Advanced · 2022 · 28 Aug 2022 · Paper 1 · Kinematics

A projectile is fired from horizontal ground with speed

v

and projection angle

θ

. When the acceleration due to gravity is

g

, the range of the projectile is

d

. If at the highest point in its trajectory, the projectile enters a different region where the effective acceleration due to gravity is

g^{'} = \frac{g}{0.81}

, then the new range is

d^{'} = n d

. The value of

n

is ____.

Answer: 0.95

Solution

Original range:

d = \frac{v ^{2} s i n 2 θ}{g}

At the highest point, the projectile has:

Time to fall from maximum height with

g^{'} = g /0.81

\frac{1}{2} g^{'} t^{2} = H \Rightarrow t = \frac{2 H}{g ^{'}} = \frac{2 v ^{2} sin ^{2} θ}{2 g \times g /0.81} = \frac{0.81 v ^{2} sin ^{2} θ}{g ^{2}} \times g = 0.9 \frac{v sin θ}{g}

Additional horizontal distance:

d_{1} = v cos θ \times t = 0.9 \frac{v ^{2} s i n 2 θ}{2 g}

New range:

d^{'} = \frac{d}{2} + d_{1} = \frac{d}{2} + 0.9 \frac{d}{2} = 0.95 d

Therefore

n = 0.95

JEE Advanced · 2022 · 28 Aug 2022 · Paper 1 · Numerical answer