JEE PYQ: Motion in a Plane - Question ID 0358eca4a0a6 (JEE Main 2024)

In projectile motion on a horizontal plane, two key quantities are:

• Horizontal range $R = \frac{u^2 \sin 2\theta}{g}$ where $u$ is the initial speed, $\theta$ is the projection angle, and $g$ is the acceleration due to gravity.
• Maximum height $H = \frac{u^2 \sin^2 \theta}{2g}$

The problem asks for the angle $\theta$ at which the horizontal range $R$ equals the maximum height $H$.

1. Write the condition $R = H$: $\frac{u^2 \sin 2\theta}{g} = \frac{u^2 \sin^2 \theta}{2g}$ 2. Cancel the common factor $\frac{u^2}{g}$ from both sides: $\sin 2\theta = \frac{\sin^2 \theta}{2}$ 3. Use the double-angle identity $\sin 2\theta = 2\sin\theta\cos\theta$: $2\sin\theta\cos\theta = \frac{\sin^2 \theta}{2}$ 4. Multiply both sides by $2$ to clear the fraction: $4\sin\theta\cos\theta = \sin^2 \theta$ 5. Divide both sides by $\sin\theta$ (assuming $\sin\theta \neq 0$): $4\cos\theta = \sin\theta$ 6. Divide both sides by $\cos\theta$ to form $\tan\theta$: $4 = \tan\theta$ 7. Take the inverse tangent of both sides: $\theta = \tan^{-1}(4)$

• Students often confuse the double-angle identity and write $\sin 2\theta$ incorrectly, leading to wrong values of $\tan\theta$. • Another frequent error is forgetting to cancel $\sin\theta$, which leaves an extra factor and gives $\tan^{-1}(1/4)$ instead of $\tan^{-1}(4)$. • Always verify the final expression by plugging $\theta = \tan^{-1}(4)$ back into the original formulas for $R$ and $H$ to confirm they are equal.