A ball is thrown from the location $\left(\mathrm{x}_{0}, \mathrm{y}_{0}\right)=(0,0)$ of a horizontal playground with an initial speed $v_{0}$ at an angle $\theta_{0}$ from the +x-direction. The ball is to be hit by a stone, which is thrown at the same time from the location $\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)=(L, 0)$. The stone is thrown at an angle $\left(180-\theta_{1}\right)$ from the +x-direction with a suitable initial speed. For a fixed $v_{0}$, when $\left(\theta_{0}, \theta_{1}\right)=\left(45^{\circ}, 45^{\circ}\right)$, the stone hits the ball after time $T_{1}$, and when $\left(\theta_{0}, \theta_{1}\right)=\left(60^{\circ}, 30^{\circ}\right)$, it hits the ball after time $T_{2}$. In such a case, $\left(T_{1} / T_{2}\right)^{2}$ is ____.

JEE Advanced · 2024 · 26 May 2024 · Paper 2 · Kinematics

A ball is thrown from the location

(x_{0}, y_{0}) = (0, 0)

of a horizontal playground with an initial speed

v_{0}

at an angle

θ_{0}

from the +x-direction. The ball is to be hit by a stone, which is thrown at the same time from the location

(x_{1}, y_{1}) = (L, 0)

. The stone is thrown at an angle

(180 - θ_{1})

from the +x-direction with a suitable initial speed. For a fixed

v_{0}

, when

(θ_{0}, θ_{1}) = (4 5^{\circ}, 4 5^{\circ})

, the stone hits the ball after time

T_{1}

, and when

(θ_{0}, θ_{1}) = (6 0^{\circ}, 3 0^{\circ})

, it hits the ball after time

T_{2}

. In such a case,

(T_{1} / T_{2})^{2}

is ____.

Answer: 2

For collision to occur, the relative acceleration in the horizontal direction must be zero (both have same vertical acceleration

g

).

Case I:

(θ_{0}, θ_{1}) = (45°, 45°)

Horizontal components: ball has

v_{0} / 2

, stone has

v / 2

(leftward).
For collision:

v = v_{0}

.

Time to meet:

T_{1} = \frac{L}{v _{0} / 2 + v _{0} / 2} = \frac{L}{2 v _{0}}

Case II:

(θ_{0}, θ_{1}) = (60°, 30°)

Horizontal components: ball has

v_{0} /2

, stone has

v 3 /2

(leftward).
For collision (matching vertical components):

v = 3 v_{0}

.

Time to meet:

T_{2} = \frac{L}{v _{0} /2 + 3 v _{0} /2} = \frac{L}{2 v _{0}}

Therefore:

(\frac{T _{1}}{T _{2}})^{2} = (\frac{L / ( 2 v _{0} )}{L / ( 2 v _{0} )})^{2} = (\frac{2}{2})^{2} = 2

JEE Advanced · 2024 · 26 May 2024 · Paper 2 · Integer answer