JEE PYQ: Motion in a Plane - Question ID 31f017e8fa2e (JEE Main 2004)

Core Formula & Concept:

In projectile motion, an object is launched at an angle $\theta$ with the horizontal and moves under the influence of gravity. Two key quantities are:

• Range ($R$): The horizontal distance traveled by the projectile before returning to the same vertical level. The formula for range is: $R = \frac{u^2 \sin 2\theta}{g}$ where $u$ is the initial speed, $\theta$ is the projection angle, and $g$ is the acceleration due to gravity.
• Time of Flight ($T$): The total time the projectile remains in the air. It is given by: $T = \frac{2u \sin \theta}{g}$

A crucial observation is that the range $R$ depends on $\sin 2\theta$, which is symmetric about $\theta = 45^\circ$. This means that for a given initial speed $u$, two different angles $\theta$ and $(90^\circ - \theta)$ yield the same range. These are called complementary angles.

Let $\theta_1 = \theta$ and $\theta_2 = 90^\circ - \theta$. Then: $\sin 2\theta_1 = \sin 2\theta, \quad \sin 2\theta_2 = \sin(180^\circ - 2\theta) = \sin 2\theta$ Hence, $R$ is the same for both angles.

The question asks about the product of the two times of flight, $T_1$ and $T_2$, corresponding to these two angles, and how it relates to $R$.

Step-by-Step Derivation:

Let’s denote:

- $T_1$ = time of flight for angle $\theta$ - $T_2$ = time of flight for angle $(90^\circ - \theta)$ Using the time of flight formula: $T_1 = \frac{2u \sin \theta}{g}, \quad T_2 = \frac{2u \sin(90^\circ - \theta)}{g} = \frac{2u \cos \theta}{g}$ Now, compute the product $T_1 \cdot T_2$: $T_1 \cdot T_2 = \left(\frac{2u \sin \theta}{g}\right) \left(\frac{2u \cos \theta}{g}\right) = \frac{4u^2 \sin \theta \cos \theta}{g^2}$ Recall the double-angle identity: $\sin 2\theta = 2 \sin \theta \cos \theta \Rightarrow \sin \theta \cos \theta = \frac{\sin 2\theta}{2}$ Substitute this into the product: $T_1 \cdot T_2 = \frac{4u^2}{g^2} \cdot \frac{\sin 2\theta}{2} = \frac{2u^2 \sin 2\theta}{g^2}$ But from the range formula: $R = \frac{u^2 \sin 2\theta}{g} \Rightarrow \sin 2\theta = \frac{Rg}{u^2}$ Substitute $\sin 2\theta$ back into the product: $T_1 \cdot T_2 = \frac{2u^2}{g^2} \cdot \frac{Rg}{u^2} = \frac{2R}{g}$ Thus: $T_1 \cdot T_2 = \frac{2}{g} R$ This shows that the product of the two times of flight is directly proportional to $R$, with the constant of proportionality being $\frac{2}{g}$.

Hence, the correct option is A: $R$. Common Traps & Exam Tip:

Students often make the following mistakes:

• Ignoring the complementary angle relationship: They may try to compute $T_1$ and $T_2$ for arbitrary angles, not realizing that the two angles are complementary ($90^\circ - \theta$). This leads to unnecessary complexity.
• Misapplying trigonometric identities: Forgetting that $\sin(90^\circ - \theta) = \cos \theta$ or not using $\sin 2\theta = 2 \sin \theta \cos \theta$ can result in incorrect expressions.
• Confusing proportionality with equality: The question asks for proportionality, not exact equality. Students may overlook the constant factor $\frac{2}{g}$ and think the product equals $R$, which is conceptually correct in terms of proportionality but may cause confusion in numerical problems.
• Assuming $T_1 \cdot T_2$ depends on $R^2$ or $1/R$: Without derivation, some guess that the product might involve $R^2$ or inverse relations, especially if they confuse time with velocity or acceleration.

Exam Tip: Always start by writing down known formulas and use trigonometric identities early. Recognize symmetry in projectile motion—complementary angles give the same range but different times of flight. This symmetry is a powerful tool in solving such problems efficiently.