Question

Two bodies A and B are projected from the same place in same vertical plane with velocities v1 and v2 from a from a long inclined
plane as shown. The ratio of their time of flight is
A) v1 sin theta / v2
B) 2v1 sin theta / v2
C) v1 sin theta / 2v2
D) v1 cos theta / v2
​

Answer 1

ANSWER:

Option A) is correct.

$\sf \dfrac{T_1}{T_2}=\dfrac{v_1 \ sin\ \theta}{ v_{2} }$

GIVEN:

• Two bodies A and B are projected from the same vertical plane.

• Their velocities are v₁ and v₂ respectively.

TO FIND:

• The ratio of their time of flight(v₁ : v₂).

EXPLANATION:

$\sf Generally \ \ \boxed{ \bold { \large{ \gray{ T = \dfrac{2u \ sin \ \theta}{g}}}}}$

But here due to inclination

$\boxed{ \bold { \large{ \gray{ T = \dfrac{2u \ sin \ \alpha}{g \ sin \ \beta}}}}}$

$\sf \beta\leadsto Angle\ of\ inclination$

$\sf\alpha \leadsto Angle\ of \ projection$

$\sf\mapsto T_1 = \dfrac{2 v_{1} \ sin \ \alpha}{g \ sin \ \beta}$

$\sf\mapsto Here \ \beta = \alpha = \theta$

As the two angles are alternate angles.

$\sf\mapsto T_1 = \dfrac{2 v_{1} \ sin \ \theta}{g \ sin \ \theta}$

$\sf\mapsto T_1 = \dfrac{2 v_{1}}{g}$

$\sf\mapsto T_2 = \dfrac{2 v_{2} \ sin \ \alpha}{g \ sin \ \beta}$

$\sf\mapsto Here \ \beta = \theta\ and \ \alpha=90^{\circ}$

$\sf\mapsto T_2= \dfrac{2 v_{2} \ sin \ 90^{\circ}}{g \ sin \ \theta}$

$\sf \mapsto T_2= \dfrac{2 v_{2} }{g \ sin \ \theta}$

$\sf\mapsto\dfrac{T_1}{T_2}=\dfrac{\dfrac{2 v_{1}}{g}}{\dfrac{2 v_{2} }{g \ sin \ \theta}}$

$\sf\mapsto\dfrac{T_1}{T_2}=\dfrac{v_1}{\dfrac{ v_{2} }{ \ sin \ \theta}}$

$\sf \mapsto\dfrac{T_1}{T_2}=\dfrac{v_1 \ sin\ \theta}{ v_{2} }$

$\bf The\ ratio\ of\ time\ of\ flight= \dfrac{v_1 \ sin\ \theta}{ v_{2} }$

Note : Refer Attachment For Diagram.