JEE PYQ: Motion in a Plane - Question ID f75f91ba4882 (JEE Main 2025)

Core Formula & Concept:

This problem deals with the fundamental principles of projectile motion, specifically focusing on the time of flight for a projectile launched from the ground. When a projectile is launched with an initial speed $u$ at an angle $\theta$ with the horizontal, its motion can be decomposed into independent horizontal and vertical components. The time of flight is solely determined by the vertical motion.

The vertical component of the initial velocity is $u_y = u \sin \theta$. Due to gravity, the vertical velocity changes, but the horizontal velocity remains constant (neglecting air resistance). The projectile starts from the ground, goes up to its maximum height, and then falls back to the ground. The total time taken for this journey is the time of flight.

Using the kinematic equation for vertical displacement, $y = u_y t + \frac{1}{2} a_y t^2$, where $y=0$ for the full flight (starting and ending at the same horizontal level), $u_y = u \sin \theta$, and $a_y = -g$ (taking upward direction as positive):

$0 = (u \sin \theta) T + \frac{1}{2} (-g) T^2$ $0 = T \left( u \sin \theta - \frac{1}{2} g T \right)$

This gives two solutions for $T$: $T=0$ (the initial moment of launch) or $u \sin \theta - \frac{1}{2} g T = 0$.

Solving for the non-zero time of flight, $T$:

$\frac{1}{2} g T = u \sin \theta$ $T = \frac{2u \sin \theta}{g}$

This is the core formula for the time of flight of a projectile launched from ground level and landing back on ground level.

Step-by-Step Derivation:

Let the initial speed of both projectiles be $u$.

For the first projectile, the angle of projection is $\theta_1 = 45^{\circ} + \alpha$. Its time of flight, $T_1$, will be:

$T_1 = \frac{2u \sin(45^{\circ} + \alpha)}{g} \quad \cdots (1)$

For the second projectile, the angle of projection is $\theta_2 = 45^{\circ} - \alpha$. Its time of flight, $T_2$, will be:

$T_2 = \frac{2u \sin(45^{\circ} - \alpha)}{g} \quad \cdots (2)$

We need to find the ratio of their times of flights, $T_1 / T_2$. Dividing equation (1) by equation (2):

$\frac{T_1}{T_2} = \frac{\frac{2u \sin(45^{\circ} + \alpha)}{g}}{\frac{2u \sin(45^{\circ} - \alpha)}{g}}$

The terms $2u$ and $g$ cancel out, simplifying the ratio to:

$\frac{T_1}{T_2} = \frac{\sin(45^{\circ} + \alpha)}{\sin(45^{\circ} - \alpha)}$

Now, we use the trigonometric sum and difference formulas for sine:

$\sin(A+B) = \sin A \cos B + \cos A \sin B$

$\sin(A-B) = \sin A \cos B - \cos A \sin B$

Applying these with $A = 45^{\circ}$ and $B = \alpha$:

$\sin(45^{\circ} + \alpha) = \sin 45^{\circ} \cos \alpha + \cos 45^{\circ} \sin \alpha$

$\sin(45^{\circ} - \alpha) = \sin 45^{\circ} \cos \alpha - \cos 45^{\circ} \sin \alpha$

We know that $\sin 45^{\circ} = \frac{1}{\sqrt{2}}$ and $\cos 45^{\circ} = \frac{1}{\sqrt{2}}$. Substituting these values:

$\sin(45^{\circ} + \alpha) = \frac{1}{\sqrt{2}} \cos \alpha + \frac{1}{\sqrt{2}} \sin \alpha = \frac{1}{\sqrt{2}} (\cos \alpha + \sin \alpha)$

$\sin(45^{\circ} - \alpha) = \frac{1}{\sqrt{2}} \cos \alpha - \frac{1}{\sqrt{2}} \sin \alpha = \frac{1}{\sqrt{2}} (\cos \alpha - \sin \alpha)$

Substitute these expressions back into the ratio for $T_1/T_2$:

$\frac{T_1}{T_2} = \frac{\frac{1}{\sqrt{2}} (\cos \alpha + \sin \alpha)}{\frac{1}{\sqrt{2}} (\cos \alpha - \sin \alpha)} = \frac{\cos \alpha + \sin \alpha}{\cos \alpha - \sin \alpha}$

To express this in terms of $\tan \alpha$, divide both the numerator and the denominator by $\cos \alpha$ (assuming $\cos \alpha \neq 0$):

$\frac{T_1}{T_2} = \frac{\frac{\cos \alpha}{\cos \alpha} + \frac{\sin \alpha}{\cos \alpha}}{\frac{\cos \alpha}{\cos \alpha} - \frac{\sin \alpha}{\cos \alpha}} = \frac{1 + \tan \alpha}{1 - \tan \alpha}$

This matches Option A.

Common Traps & Exam Tip:

Common Traps:

1. Trigonometric Errors: Incorrectly applying sum/difference formulas for sine or making calculation mistakes with $\sin 45^{\circ}$ and $\cos 45^{\circ}$. A common mistake is to write $\sin(45^{\circ}+\alpha) = \sin 45^{\circ} + \sin \alpha$, which is incorrect.
2. Algebraic Manipulation Errors: While simplifying the ratio $\frac{\cos \alpha + \sin \alpha}{\cos \alpha - \sin \alpha}$, students might incorrectly divide only one term or make sign errors.
3. Confusing with Range Formula: Sometimes students confuse the conditions for maximum range or equal ranges. For range, angles $\theta$ and $(90^\circ - \theta)$ give the same range for a given initial speed. However, for time of flight, this is not true.

Exam Tip (Alternative Approach / Quicker Method):

Observe the given angles: $\theta_1 = 45^{\circ} + \alpha$ and $\theta_2 = 45^{\circ} - \alpha$.

Notice that $\theta_1 + \theta_2 = (45^{\circ} + \alpha) + (45^{\circ} - \alpha) = 90^{\circ}$. This means the two angles of projection are complementary.

So, $\theta_2 = 90^{\circ} - \theta_1$.

Therefore, $\sin \theta_2 = \sin(90^{\circ} - \theta_1) = \cos \theta_1$.

The ratio of times of flight is $\frac{T_1}{T_2} = \frac{\sin \theta_1}{\sin \theta_2}$.

Substituting $\sin \theta_2 = \cos \theta_1$:

$\frac{T_1}{T_2} = \frac{\sin \theta_1}{\cos \theta_1} = \tan \theta_1$

Now, substitute $\theta_1 = 45^{\circ} + \alpha$:

$\frac{T_1}{T_2} = \tan(45^{\circ} + \alpha)$

Using the tangent addition formula: $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$.

Here, $A = 45^{\circ}$ and $B = \alpha$. Since $\tan 45^{\circ} = 1$:

$\tan(45^{\circ} + \alpha) = \frac{\tan 45^{\circ} + \tan \alpha}{1 - \tan 45^{\circ} \tan \alpha} = \frac{1 + \tan \alpha}{1 - 1 \cdot \tan \alpha} = \frac{1 + \tan \alpha}{1 - \tan \alpha}$

This approach is significantly faster if you recognize the complementary angle relationship and are proficient with trigonometric identities, especially the tangent addition formula. Always look for such relationships in projectile motion problems as they often simplify calculations dramatically.