Two projectiles are fired from ground with same initial speeds from same point at angles (45° + α) and (45° - α) with horizontal direction. The ratio of their times of flights is

The correct answer is Option 1. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Physics Q31 – JEE-MAIN 2025

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Home Physics Motion in a PlaneQuestion #311

Two projectiles are fired from ground with same initial speeds from same point at angles

(45^{\circ} + α)

and

(45^{\circ} - α)

with horizontal direction. The ratio of their times of flights is

Clue 1 of 3: Projectile Motion Fundamentals

Recall the formula for the time of flight of a projectile launched from the ground.

🥷

Ninja Strategy•Limiting Case Analysis

Test the options with a special value of $α$ . For $α = 45^{\circ}$ , the angles are $90^{\circ}$ and $0^{\circ}$ . The time of flight for $0^{\circ}$ is 0, making the ratio undefined. Only option 1 yields an undefined result for $α = 45^{\circ}$ (due to division by zero).

🧠 Full Solution Path

Step 1: Identify Time of Flight Formula✦ Active

The time of flight for a projectile launched from the ground with initial speed $u$ at an angle $θ$ with the horizontal is given by:

T = \frac{2 u \sin θ}{g}

Step 2: Formulate Ratio of Times○ Expand

For the first projectile, $θ_{1} = 45^{\circ} + α$ , and for the second, $θ_{2} = 45^{\circ} - α$ . The initial speed $u$ and acceleration due to gravity $g$ are the same for both. The ratio of their times of flight is:

\frac{T_{1}}{T_{2}} = \frac{\frac{2 u \sin (45^{\circ} + α)}{g}}{\frac{2 u \sin (45^{\circ} - α)}{g}} = \frac{\sin (45^{\circ} + α)}{\sin (45^{\circ} - α)}

Step 3: Apply Trigonometric Identities and Simplify○ Expand

Using the identity $\sin (A \pm B) = \sin A \cos B \pm \cos A \sin B$ with $A = 45^{\circ}$ and $B = α$ , and knowing $\sin 45^{\circ} = \cos 45^{\circ} = \frac{1}{\sqrt{2}}$ :

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

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

\frac{T_{1}}{T_{2}} = \frac{\frac{1}{\sqrt{2}} (\cos α + \sin α)}{\frac{1}{\sqrt{2}} (\cos α - \sin α)} = \frac{\cos α + \sin α}{\cos α - \sin α}

Divide the numerator and denominator by $\cos α$ to express the ratio in terms of $\tan α$ :

\frac{T_{1}}{T_{2}} = \frac{1 + \tan α}{1 - \tan α}

💡 Teacher's Secret Hint

Ensure $\cos α \neq 0$ and $1 - \tan α \neq 0$ for the expression to be well-defined.

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Physics Question 31 – JEE-MAIN 2025

Clue 1 of 3: Projectile Motion Fundamentals

🧠 Full Solution Path